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Set Based probability Questions in English
Class 11 Mathematics · Probability · Set Based probability
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251
Medium
Two dice are thrown. The events $A, B$ and $C$ are as follows:
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
Describe the events $A$ and $B$.
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
Describe the events $A$ and $B$.
Solution
(A) When two dice are thrown,the sample space $S$ is given by $S = \{(x, y) : x, y \in \{1, 2, 3, 4, 5, 6\}\}$.
Event $A$ is defined as getting an even number on the first die. Thus,the first coordinate $x$ must be $2, 4,$ or $6$.
$A = \{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$.
Event $B$ is defined as getting an odd number on the first die. Thus,the first coordinate $x$ must be $1, 3,$ or $5$.
$B = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$.
Event $A$ is defined as getting an even number on the first die. Thus,the first coordinate $x$ must be $2, 4,$ or $6$.
$A = \{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\}$.
Event $B$ is defined as getting an odd number on the first die. Thus,the first coordinate $x$ must be $1, 3,$ or $5$.
$B = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$.
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MediumMCQ
If a leap year is selected at random,what is the chance that it will contain $53$ Tuesdays?
A
$\frac{1}{7}$
B
$\frac{2}{7}$
C
$\frac{3}{7}$
D
$\frac{4}{7}$
Solution
(B) leap year consists of $366$ days,which is equal to $52$ weeks and $2$ extra days.
In $52$ weeks,there are exactly $52$ Tuesdays.
To have $53$ Tuesdays,one of the remaining $2$ days must be a Tuesday.
The possible pairs for the $2$ extra days are:
(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),(Saturday,Sunday),(Sunday,Monday).
Total number of possible outcomes $= 7$.
The favorable outcomes where at least one day is a Tuesday are (Monday,Tuesday) and (Tuesday,Wednesday).
Number of favorable outcomes $= 2$.
Therefore,the probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{2}{7}$.
In $52$ weeks,there are exactly $52$ Tuesdays.
To have $53$ Tuesdays,one of the remaining $2$ days must be a Tuesday.
The possible pairs for the $2$ extra days are:
(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),(Saturday,Sunday),(Sunday,Monday).
Total number of possible outcomes $= 7$.
The favorable outcomes where at least one day is a Tuesday are (Monday,Tuesday) and (Tuesday,Wednesday).
Number of favorable outcomes $= 2$.
Therefore,the probability $= \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{2}{7}$.
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View Solution253
MediumMCQ
Two dice are rolled. If both dice have six faces numbered $1, 2, 3, 5, 7,$ and $11,$ then the probability that the sum of the numbers on the top faces is less than or equal to $8$ is
A
$\frac{4}{9}$
B
$\frac{17}{36}$
C
$\frac{5}{12}$
D
$\frac{1}{2}$
Solution
(B) The total number of outcomes when two dice are rolled is $6 \times 6 = 36$.
Let the outcomes be $(x, y)$ where $x, y \in \{1, 2, 3, 5, 7, 11\}$.
We need the sum $x + y \leq 8$.
The possible pairs $(x, y)$ are:
If $x=1$,$y \in \{1, 2, 3, 5, 7\}$ ($5$ outcomes).
If $x=2$,$y \in \{1, 2, 3, 5\}$ ($4$ outcomes).
If $x=3$,$y \in \{1, 2, 3, 5\}$ ($4$ outcomes).
If $x=5$,$y \in \{1, 2, 3\}$ ($3$ outcomes).
If $x=7$,$y \in \{1\}$ ($1$ outcome).
If $x=11$,no $y$ satisfies the condition.
Total favorable outcomes $n(E) = 5 + 4 + 4 + 3 + 1 = 17$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{17}{36}$.
Let the outcomes be $(x, y)$ where $x, y \in \{1, 2, 3, 5, 7, 11\}$.
We need the sum $x + y \leq 8$.
The possible pairs $(x, y)$ are:
If $x=1$,$y \in \{1, 2, 3, 5, 7\}$ ($5$ outcomes).
If $x=2$,$y \in \{1, 2, 3, 5\}$ ($4$ outcomes).
If $x=3$,$y \in \{1, 2, 3, 5\}$ ($4$ outcomes).
If $x=5$,$y \in \{1, 2, 3\}$ ($3$ outcomes).
If $x=7$,$y \in \{1\}$ ($1$ outcome).
If $x=11$,no $y$ satisfies the condition.
Total favorable outcomes $n(E) = 5 + 4 + 4 + 3 + 1 = 17$.
Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{17}{36}$.
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View Solution254
Medium
Two dice are thrown. The events $A, B$ and $C$ are as follows:
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
Describe the events $B$ and $C$.
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
Describe the events $B$ and $C$.
Solution
(N/A) When two dice are thrown,the sample space $S$ contains $36$ outcomes:
$S = \{(x, y) : x, y \in \{1, 2, 3, 4, 5, 6\} \}$.
Event $B$ is getting an odd number on the first die:
$B = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$.
Event $C$ is getting the sum of the numbers on the dice $\leq 5$:
$C = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)\}$.
$S = \{(x, y) : x, y \in \{1, 2, 3, 4, 5, 6\} \}$.
Event $B$ is getting an odd number on the first die:
$B = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)\}$.
Event $C$ is getting the sum of the numbers on the dice $\leq 5$:
$C = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)\}$.
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View Solution255
Medium
Two dice are thrown. The events $A$,$B$,and $C$ are as follows:
$A$: getting an even number on the first die.
$B$: getting an odd number on the first die.
$C$: getting the sum of the numbers on the dice $\leq 5$.
State true or false: (give reason for your answer)
Statement: $A$ and $B$ are mutually exclusive.
$A$: getting an even number on the first die.
$B$: getting an odd number on the first die.
$C$: getting the sum of the numbers on the dice $\leq 5$.
State true or false: (give reason for your answer)
Statement: $A$ and $B$ are mutually exclusive.
Solution
(A) The sample space $S$ for throwing two dice contains $36$ outcomes.
Event $A$ (even number on the first die) is given by:
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Event $B$ (odd number on the first die) is given by:
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
Two events are mutually exclusive if their intersection is an empty set,i.e.,$A \cap B = \phi$.
Since the first die cannot be both even and odd simultaneously,there are no common outcomes between $A$ and $B$.
Thus,$A \cap B = \phi$.
Therefore,the statement is true.
Event $A$ (even number on the first die) is given by:
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Event $B$ (odd number on the first die) is given by:
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
Two events are mutually exclusive if their intersection is an empty set,i.e.,$A \cap B = \phi$.
Since the first die cannot be both even and odd simultaneously,there are no common outcomes between $A$ and $B$.
Thus,$A \cap B = \phi$.
Therefore,the statement is true.
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View Solution256
Medium
Two dice are thrown. The events $A, B$ and $C$ are as follows:
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
State true or false: (give reason for your answer)
Statement: $A$ and $B$ are mutually exclusive and exhaustive.
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
State true or false: (give reason for your answer)
Statement: $A$ and $B$ are mutually exclusive and exhaustive.
Solution
(A) The sample space $S$ for throwing two dice contains $36$ outcomes.
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
Two events are mutually exclusive if $A \cap B = \phi$. Here,$A$ contains only even numbers on the first die and $B$ contains only odd numbers on the first die,so $A \cap B = \phi$.
Two events are exhaustive if $A \cup B = S$. Since every outcome in $S$ has either an even or an odd number on the first die,$A \cup B = S$.
Therefore,the statement is true.
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
Two events are mutually exclusive if $A \cap B = \phi$. Here,$A$ contains only even numbers on the first die and $B$ contains only odd numbers on the first die,so $A \cap B = \phi$.
Two events are exhaustive if $A \cup B = S$. Since every outcome in $S$ has either an even or an odd number on the first die,$A \cup B = S$.
Therefore,the statement is true.
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View Solution257
Easy
Two dice are thrown. The events $A, B$ and $C$ are as follows:
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
State whether the following statement is true or false and provide a reason:
Statement: $A = B^{\prime}$
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
State whether the following statement is true or false and provide a reason:
Statement: $A = B^{\prime}$
Solution
(A) The sample space $S$ for throwing two dice consists of $36$ outcomes.
Event $A$ is getting an even number on the first die:
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Event $B$ is getting an odd number on the first die:
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
The complement of $B$,denoted as $B^{\prime}$,consists of all outcomes in $S$ that are not in $B$. Since the first die can only show an even or odd number,the complement of getting an odd number on the first die is getting an even number on the first die.
Therefore,$B^{\prime} = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Comparing the sets,we see that $A = B^{\prime}$.
Thus,the statement is true.
Event $A$ is getting an even number on the first die:
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Event $B$ is getting an odd number on the first die:
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
The complement of $B$,denoted as $B^{\prime}$,consists of all outcomes in $S$ that are not in $B$. Since the first die can only show an even or odd number,the complement of getting an odd number on the first die is getting an even number on the first die.
Therefore,$B^{\prime} = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
Comparing the sets,we see that $A = B^{\prime}$.
Thus,the statement is true.
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Easy
Two dice are thrown. The events $A, B$ and $C$ are as follows:
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
State true or false: (give reason for your answer)
Statement: $A', B', C$ are mutually exclusive and exhaustive.
$A:$ getting an even number on the first die.
$B:$ getting an odd number on the first die.
$C:$ getting the sum of the numbers on the dice $\leq 5$.
State true or false: (give reason for your answer)
Statement: $A', B', C$ are mutually exclusive and exhaustive.
Solution
(B) The sample space $S$ contains $36$ outcomes.
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
Since $A$ is the event of getting an even number on the first die,$A' = B$. Similarly,$B' = A$.
$C = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)\}$
For events to be mutually exclusive,the intersection of any two events must be empty $(\phi)$.
Check $B' \cap C = A \cap C = \{(2,1), (2,2), (2,3), (4,1)\} \neq \phi$.
Since the intersection is not empty,the events are not mutually exclusive.
Therefore,the statement is false.
$A = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$
Since $A$ is the event of getting an even number on the first die,$A' = B$. Similarly,$B' = A$.
$C = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)\}$
For events to be mutually exclusive,the intersection of any two events must be empty $(\phi)$.
Check $B' \cap C = A \cap C = \{(2,1), (2,2), (2,3), (4,1)\} \neq \phi$.
Since the intersection is not empty,the events are not mutually exclusive.
Therefore,the statement is false.
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View Solution259
EasyMCQ
One card is drawn from a well-shuffled deck of $52$ cards. If each outcome is equally likely,calculate the probability that the card will be a diamond.
A
$1/4$
B
$1/13$
C
$1/2$
D
$3/4$
Solution
(A) The total number of cards in a well-shuffled deck is $52$.
There are $4$ suits in a deck of cards,and each suit contains $13$ cards.
The number of diamond cards is $13$.
The probability $P(A)$ of drawing a diamond card is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$P(A) = \frac{13}{52} = \frac{1}{4}$.
There are $4$ suits in a deck of cards,and each suit contains $13$ cards.
The number of diamond cards is $13$.
The probability $P(A)$ of drawing a diamond card is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$P(A) = \frac{13}{52} = \frac{1}{4}$.
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View Solution260
EasyMCQ
One card is drawn from a well-shuffled deck of $52$ cards. If each outcome is equally likely,calculate the probability that the card will be a diamond but not an ace.
A
$\frac{3}{52}$
B
$\frac{1}{4}$
C
$\frac{12}{52}$
D
$\frac{1}{13}$
Solution
(C) Total number of cards in a deck = $52$.
Number of diamond cards = $13$.
There is $1$ ace of diamonds in the deck.
We need to find the probability of drawing a diamond that is not an ace.
Number of favorable outcomes = (Total diamond cards) - (Ace of diamonds) = $13 - 1 = 12$.
Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{52} = \frac{3}{13}$.
Number of diamond cards = $13$.
There is $1$ ace of diamonds in the deck.
We need to find the probability of drawing a diamond that is not an ace.
Number of favorable outcomes = (Total diamond cards) - (Ace of diamonds) = $13 - 1 = 12$.
Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{52} = \frac{3}{13}$.
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View Solution261
EasyMCQ
One card is drawn from a well-shuffled deck of $52$ cards. If each outcome is equally likely,calculate the probability that the card will be a black card (i.e.,a club or a spade).
A
$1/2$
B
$1/4$
C
$1/13$
D
$3/4$
Solution
(A) The total number of cards in a deck is $52$. So,the total number of possible outcomes is $52$.
$A$ standard deck contains two black suits: clubs and spades. Each suit has $13$ cards.
Therefore,the total number of black cards is $13 + 13 = 26$.
Let $C$ be the event of drawing a black card. The number of favorable outcomes is $26$.
The probability $P(C)$ is given by the ratio of favorable outcomes to total outcomes:
$P(C) = \frac{26}{52} = \frac{1}{2}$.
$A$ standard deck contains two black suits: clubs and spades. Each suit has $13$ cards.
Therefore,the total number of black cards is $13 + 13 = 26$.
Let $C$ be the event of drawing a black card. The number of favorable outcomes is $26$.
The probability $P(C)$ is given by the ratio of favorable outcomes to total outcomes:
$P(C) = \frac{26}{52} = \frac{1}{2}$.
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View Solution262
EasyMCQ
One card is drawn from a well-shuffled deck of $52$ cards. If each outcome is equally likely,calculate the probability that the card will not be a diamond.
A
$\frac{1}{4}$
B
$\frac{1}{2}$
C
$\frac{3}{4}$
D
$\frac{1}{13}$
Solution
(C) The total number of cards in a well-shuffled deck is $52$.
Let $A$ be the event that the card drawn is a diamond.
There are $13$ diamond cards in a deck of $52$ cards.
Therefore,the probability of drawing a diamond is $P(A) = \frac{13}{52} = \frac{1}{4}$.
The event 'the card drawn is not a diamond' is the complement of event $A$,denoted as $A'$.
The probability of the complement event is given by $P(A') = 1 - P(A)$.
Substituting the value,we get $P(A') = 1 - \frac{1}{4} = \frac{3}{4}$.
Let $A$ be the event that the card drawn is a diamond.
There are $13$ diamond cards in a deck of $52$ cards.
Therefore,the probability of drawing a diamond is $P(A) = \frac{13}{52} = \frac{1}{4}$.
The event 'the card drawn is not a diamond' is the complement of event $A$,denoted as $A'$.
The probability of the complement event is given by $P(A') = 1 - P(A)$.
Substituting the value,we get $P(A') = 1 - \frac{1}{4} = \frac{3}{4}$.
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View Solution263
EasyMCQ
One card is drawn from a well-shuffled deck of $52$ cards. If each outcome is equally likely,calculate the probability that the card will not be a black card.
A
$1/4$
B
$1/2$
C
$3/4$
D
$1/13$
Solution
(B) Total number of cards in a deck $= 52$.
Number of black cards (spades and clubs) $= 26$.
Number of red cards (hearts and diamonds) $= 26$.
Let $E$ be the event that the card drawn is not a black card. This means the card must be a red card.
Number of favorable outcomes $= 26$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{26}{52} = \frac{1}{2}$.
Number of black cards (spades and clubs) $= 26$.
Number of red cards (hearts and diamonds) $= 26$.
Let $E$ be the event that the card drawn is not a black card. This means the card must be a red card.
Number of favorable outcomes $= 26$.
Probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{26}{52} = \frac{1}{2}$.
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View Solution264
EasyMCQ
$A$ bag contains $9$ discs of which $4$ are red,$3$ are blue and $2$ are yellow. The discs are similar in shape and size. $A$ disc is drawn at random from the bag. Calculate the probability that it will be red. (in $/9$)
A
$4$
B
$3$
C
$2$
D
$1$
Solution
(A) The total number of discs in the bag is $9$. Therefore,the total number of possible outcomes is $9$.
Let $A$ be the event that the disc drawn is red.
The number of red discs is $4$,so the number of favorable outcomes is $n(A) = 4$.
The probability of drawing a red disc is given by the formula:
$P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$
$P(A) = \frac{4}{9}$
Let $A$ be the event that the disc drawn is red.
The number of red discs is $4$,so the number of favorable outcomes is $n(A) = 4$.
The probability of drawing a red disc is given by the formula:
$P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$
$P(A) = \frac{4}{9}$
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EasyMCQ
$A$ bag contains $9$ discs of which $4$ are red,$3$ are blue and $2$ are yellow. The discs are similar in shape and size. $A$ disc is drawn at random from the bag. Calculate the probability that it will be yellow. (in $/9$)
A
$2$
B
$4$
C
$3$
D
$1$
Solution
(A) Total number of discs $= 4 + 3 + 2 = 9$.
So,the total number of possible outcomes is $9$.
Let $E$ be the event of drawing a yellow disc.
The number of yellow discs is $2$.
Therefore,the number of favorable outcomes is $2$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{9}$.
So,the total number of possible outcomes is $9$.
Let $E$ be the event of drawing a yellow disc.
The number of yellow discs is $2$.
Therefore,the number of favorable outcomes is $2$.
The probability $P(E)$ is given by the formula:
$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{2}{9}$.
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View Solution266
EasyMCQ
$A$ bag contains $9$ discs of which $4$ are red,$3$ are blue,and $2$ are yellow. The discs are similar in shape and size. $A$ disc is drawn at random from the bag. Calculate the probability that it will be blue.
A
$1/9$
B
$1/3$
C
$2/9$
D
$4/9$
Solution
(B) The total number of discs in the bag is $9$,so the total number of possible outcomes is $9$.
Let $C$ be the event that the disc drawn is blue.
The number of blue discs is $3$,so the number of favorable outcomes is $n(C) = 3$.
The probability of drawing a blue disc is given by $P(C) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
$P(C) = \frac{3}{9} = \frac{1}{3}$.
Let $C$ be the event that the disc drawn is blue.
The number of blue discs is $3$,so the number of favorable outcomes is $n(C) = 3$.
The probability of drawing a blue disc is given by $P(C) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
$P(C) = \frac{3}{9} = \frac{1}{3}$.
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EasyMCQ
$A$ bag contains $9$ discs of which $4$ are red,$3$ are blue and $2$ are yellow. The discs are similar in shape and size. $A$ disc is drawn at random from the bag. Calculate the probability that it will be not blue.
A
$\frac{1}{3}$
B
$\frac{2}{3}$
C
$\frac{4}{9}$
D
$\frac{5}{9}$
Solution
(B) The total number of discs in the bag is $9$. So,the total number of possible outcomes is $9$.
Let $C$ be the event that the disc drawn is blue.
The number of blue discs is $3$.
Therefore,the probability of drawing a blue disc is $P(C) = \frac{3}{9} = \frac{1}{3}$.
The event 'not blue' is the complement of event $C$,denoted as $C'$.
We know that $P(C') = 1 - P(C)$.
Thus,$P(C') = 1 - \frac{1}{3} = \frac{2}{3}$.
Let $C$ be the event that the disc drawn is blue.
The number of blue discs is $3$.
Therefore,the probability of drawing a blue disc is $P(C) = \frac{3}{9} = \frac{1}{3}$.
The event 'not blue' is the complement of event $C$,denoted as $C'$.
We know that $P(C') = 1 - P(C)$.
Thus,$P(C') = 1 - \frac{1}{3} = \frac{2}{3}$.
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View Solution268
MediumMCQ
$A$ bag contains $9$ discs of which $4$ are red,$3$ are blue,and $2$ are yellow. The discs are similar in shape and size. $A$ disc is drawn at random from the bag. Calculate the probability that it will be either red or blue.
A
$\frac{1}{9}$
B
$\frac{2}{9}$
C
$\frac{7}{9}$
D
$\frac{4}{9}$
Solution
(C) Total number of discs $= 4 + 3 + 2 = 9$.
Total number of possible outcomes $= 9$.
Let $R$ be the event of drawing a red disc and $B$ be the event of drawing a blue disc.
Number of red discs $= 4$,so $P(R) = \frac{4}{9}$.
Number of blue discs $= 3$,so $P(B) = \frac{3}{9}$.
Since the events are mutually exclusive,the probability of drawing either red or blue is $P(R \cup B) = P(R) + P(B)$.
$P(R \cup B) = \frac{4}{9} + \frac{3}{9} = \frac{7}{9}$.
Total number of possible outcomes $= 9$.
Let $R$ be the event of drawing a red disc and $B$ be the event of drawing a blue disc.
Number of red discs $= 4$,so $P(R) = \frac{4}{9}$.
Number of blue discs $= 3$,so $P(B) = \frac{3}{9}$.
Since the events are mutually exclusive,the probability of drawing either red or blue is $P(R \cup B) = P(R) + P(B)$.
$P(R \cup B) = \frac{4}{9} + \frac{3}{9} = \frac{7}{9}$.
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View Solution269
MediumMCQ
Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is $0.05$ and that Ashima will qualify the examination is $0.10$. The probability that both will qualify the examination is $0.02$. Find the probability that both Anil and Ashima will not qualify the examination.
A
$0.87$
B
$0.13$
C
$0.15$
D
$0.98$
Solution
(A) Let $E$ and $F$ denote the events that Anil and Ashima will qualify the examination,respectively.
Given that $P(E) = 0.05$,$P(F) = 0.10$,and $P(E \cap F) = 0.02$.
The event 'both Anil and Ashima will not qualify the examination' is represented as $E' \cap F'$.
By De Morgan's Law,$E' \cap F' = (E \cup F)'$.
First,we calculate the probability that at least one of them qualifies:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
$P(E \cup F) = 0.05 + 0.10 - 0.02 = 0.13$.
Now,the probability that neither qualifies is:
$P(E' \cap F') = P((E \cup F)') = 1 - P(E \cup F)$
$P(E' \cap F') = 1 - 0.13 = 0.87$.
Given that $P(E) = 0.05$,$P(F) = 0.10$,and $P(E \cap F) = 0.02$.
The event 'both Anil and Ashima will not qualify the examination' is represented as $E' \cap F'$.
By De Morgan's Law,$E' \cap F' = (E \cup F)'$.
First,we calculate the probability that at least one of them qualifies:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
$P(E \cup F) = 0.05 + 0.10 - 0.02 = 0.13$.
Now,the probability that neither qualifies is:
$P(E' \cap F') = P((E \cup F)') = 1 - P(E \cup F)$
$P(E' \cap F') = 1 - 0.13 = 0.87$.
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EasyMCQ
Two students,Anil and Ashima,appeared in an examination. The probability that Anil will qualify the examination is $0.05$ and that Ashima will qualify the examination is $0.10$. The probability that both will qualify the examination is $0.02$. Find the probability that at least one of them will not qualify the examination.
A
$0.98$
B
$0.95$
C
$0.90$
D
$0.87$
Solution
(A) Let $E$ and $F$ be the events that Anil and Ashima qualify the examination,respectively.
Given:
$P(E) = 0.05$
$P(F) = 0.10$
$P(E \cap F) = 0.02$
We need to find the probability that at least one of them will not qualify the examination.
This is the complement of the event that both of them qualify the examination.
Let $A$ be the event that both qualify,i.e.,$A = E \cap F$.
The probability that at least one of them will not qualify is $1 - P(E \cap F)$.
$= 1 - 0.02$
$= 0.98$
Given:
$P(E) = 0.05$
$P(F) = 0.10$
$P(E \cap F) = 0.02$
We need to find the probability that at least one of them will not qualify the examination.
This is the complement of the event that both of them qualify the examination.
Let $A$ be the event that both qualify,i.e.,$A = E \cap F$.
The probability that at least one of them will not qualify is $1 - P(E \cap F)$.
$= 1 - 0.02$
$= 0.98$
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MediumMCQ
Two students,Anil and Ashima,appeared in an examination. The probability that Anil will qualify the examination is $0.05$ and that Ashima will qualify the examination is $0.10$. The probability that both will qualify the examination is $0.02$. Find the probability that only one of them will qualify the examination.
A
$0.11$
B
$0.13$
C
$0.15$
D
$0.07$
Solution
(A) Let $E$ and $F$ denote the events that Anil and Ashima will qualify the examination,respectively.
Given that $P(E) = 0.05$,$P(F) = 0.10$,and $P(E \cap F) = 0.02$.
The event that only one of them will qualify the examination is given by $(E \cap F') \cup (E' \cap F)$,where $E'$ and $F'$ are the complements of $E$ and $F$ respectively.
Since these events are mutually exclusive,the probability is:
$P(\text{only one}) = P(E \cap F') + P(E' \cap F)$
$= [P(E) - P(E \cap F)] + [P(F) - P(E \cap F)]$
$= (0.05 - 0.02) + (0.10 - 0.02)$
$= 0.03 + 0.08 = 0.11$.
Given that $P(E) = 0.05$,$P(F) = 0.10$,and $P(E \cap F) = 0.02$.
The event that only one of them will qualify the examination is given by $(E \cap F') \cup (E' \cap F)$,where $E'$ and $F'$ are the complements of $E$ and $F$ respectively.
Since these events are mutually exclusive,the probability is:
$P(\text{only one}) = P(E \cap F') + P(E' \cap F)$
$= [P(E) - P(E \cap F)] + [P(F) - P(E \cap F)]$
$= (0.05 - 0.02) + (0.10 - 0.02)$
$= 0.03 + 0.08 = 0.11$.
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Easy
Which of the following cannot be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$?
| Outcome | Probability |
| $\omega_{1}$ | $\frac{1}{14}$ |
| $\omega_{2}$ | $\frac{2}{14}$ |
| $\omega_{3}$ | $\frac{3}{14}$ |
| $\omega_{4}$ | $\frac{4}{14}$ |
| $\omega_{5}$ | $\frac{5}{14}$ |
| $\omega_{6}$ | $\frac{6}{14}$ |
| $\omega_{7}$ | $\frac{15}{14}$ |
Solution
(D) For any probability assignment to be valid,it must satisfy two conditions:
$1$. Each probability $P(\omega_{i})$ must be such that $0 \leq P(\omega_{i}) \leq 1$.
$2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum P(\omega_{i}) = 1$.
In the given table,we observe the probability assigned to $\omega_{7}$ is $P(\omega_{7}) = \frac{15}{14}$.
Since $\frac{15}{14} > 1$,this violates the fundamental axiom of probability that states $P(\omega_{i}) \leq 1$ for all $i$.
Therefore,this assignment is not valid.
$1$. Each probability $P(\omega_{i})$ must be such that $0 \leq P(\omega_{i}) \leq 1$.
$2$. The sum of all probabilities must be equal to $1$,i.e.,$\sum P(\omega_{i}) = 1$.
In the given table,we observe the probability assigned to $\omega_{7}$ is $P(\omega_{7}) = \frac{15}{14}$.
Since $\frac{15}{14} > 1$,this violates the fundamental axiom of probability that states $P(\omega_{i}) \leq 1$ for all $i$.
Therefore,this assignment is not valid.
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EasyMCQ
$A$ coin is tossed twice. What is the probability that at least one tail occurs?
A
$1/4$
B
$1/2$
C
$3/4$
D
$1$
Solution
(C) When a coin is tossed twice,the sample space $S$ is given by:
$S = \{HH, HT, TH, TT\}$
Total number of outcomes $n(S) = 4$.
Let $A$ be the event of the occurrence of at least one tail.
The favorable outcomes are $HT, TH,$ and $TT$.
So,$A = \{HT, TH, TT\}$ and $n(A) = 3$.
The probability $P(A)$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{3}{4}$
$S = \{HH, HT, TH, TT\}$
Total number of outcomes $n(S) = 4$.
Let $A$ be the event of the occurrence of at least one tail.
The favorable outcomes are $HT, TH,$ and $TT$.
So,$A = \{HT, TH, TT\}$ and $n(A) = 3$.
The probability $P(A)$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{3}{4}$
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EasyMCQ
$A$ die is thrown. Find the probability of the following event: $A$ prime number will appear.
A
$1/2$
B
$1/3$
C
$1/6$
D
$2/3$
Solution
(A) The sample space $S$ of the experiment of throwing a die is given by:
$S = \{1, 2, 3, 4, 5, 6\}$
Total number of possible outcomes $n(S) = 6$.
Let $A$ be the event of the occurrence of a prime number.
The prime numbers in the sample space are $2, 3,$ and $5$.
So,$A = \{2, 3, 5\}$.
Number of favourable outcomes $n(A) = 3$.
The probability of event $A$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.
$S = \{1, 2, 3, 4, 5, 6\}$
Total number of possible outcomes $n(S) = 6$.
Let $A$ be the event of the occurrence of a prime number.
The prime numbers in the sample space are $2, 3,$ and $5$.
So,$A = \{2, 3, 5\}$.
Number of favourable outcomes $n(A) = 3$.
The probability of event $A$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.
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EasyMCQ
$A$ die is thrown. Find the probability of the following event: $A$ number greater than or equal to $3$ will appear.
A
$1/3$
B
$1/2$
C
$2/3$
D
$5/6$
Solution
(C) The sample space of the given experiment is $S = \{1, 2, 3, 4, 5, 6\}$.
Let $B$ be the event of the occurrence of a number greater than or equal to $3$.
Accordingly,$B = \{3, 4, 5, 6\}$.
The number of favourable outcomes is $n(B) = 4$.
The total number of possible outcomes is $n(S) = 6$.
Therefore,the probability $P(B) = \frac{n(B)}{n(S)} = \frac{4}{6} = \frac{2}{3}$.
Let $B$ be the event of the occurrence of a number greater than or equal to $3$.
Accordingly,$B = \{3, 4, 5, 6\}$.
The number of favourable outcomes is $n(B) = 4$.
The total number of possible outcomes is $n(S) = 6$.
Therefore,the probability $P(B) = \frac{n(B)}{n(S)} = \frac{4}{6} = \frac{2}{3}$.
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EasyMCQ
$A$ die is thrown. Find the probability of the following event: $A$ number less than or equal to $1$ will appear.
A
$\frac{1}{6}$
B
$\frac{1}{3}$
C
$\frac{1}{2}$
D
$\frac{2}{3}$
Solution
(A) The sample space of the given experiment is $S = \{1, 2, 3, 4, 5, 6\}$.
Let $C$ be the event of the occurrence of a number less than or equal to $1$.
Accordingly,$C = \{1\}$.
Therefore,the number of favourable outcomes $n(C) = 1$ and the total number of possible outcomes $n(S) = 6$.
$\therefore P(C) = \frac{n(C)}{n(S)} = \frac{1}{6}$.
Let $C$ be the event of the occurrence of a number less than or equal to $1$.
Accordingly,$C = \{1\}$.
Therefore,the number of favourable outcomes $n(C) = 1$ and the total number of possible outcomes $n(S) = 6$.
$\therefore P(C) = \frac{n(C)}{n(S)} = \frac{1}{6}$.
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EasyMCQ
$A$ die is thrown. Find the probability of the following event: $A$ number more than $6$ will appear.
A
$0$
B
$1/6$
C
$1/3$
D
$1$
Solution
(A) The sample space $S$ of the experiment of throwing a die is given by $S = \{1, 2, 3, 4, 5, 6\}$.
Total number of possible outcomes $n(S) = 6$.
Let $D$ be the event of the occurrence of a number greater than $6$.
Since there is no number greater than $6$ on a standard die,$D = \emptyset$.
Therefore,the number of favorable outcomes $n(D) = 0$.
The probability $P(D)$ is given by $P(D) = \frac{n(D)}{n(S)} = \frac{0}{6} = 0$.
Total number of possible outcomes $n(S) = 6$.
Let $D$ be the event of the occurrence of a number greater than $6$.
Since there is no number greater than $6$ on a standard die,$D = \emptyset$.
Therefore,the number of favorable outcomes $n(D) = 0$.
The probability $P(D)$ is given by $P(D) = \frac{n(D)}{n(S)} = \frac{0}{6} = 0$.
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EasyMCQ
$A$ die is thrown. Find the probability of the following event: $A$ number less than $6$ will appear.
A
$1/6$
B
$2/3$
C
$5/6$
D
$1/2$
Solution
(C) The sample space of the given experiment is $S = \{1, 2, 3, 4, 5, 6\}$.
Let $E$ be the event of the occurrence of a number less than $6$.
Accordingly,$E = \{1, 2, 3, 4, 5\}$.
Therefore,the number of favorable outcomes is $n(E) = 5$ and the total number of possible outcomes is $n(S) = 6$.
$P(E) = \frac{n(E)}{n(S)} = \frac{5}{6}$.
Let $E$ be the event of the occurrence of a number less than $6$.
Accordingly,$E = \{1, 2, 3, 4, 5\}$.
Therefore,the number of favorable outcomes is $n(E) = 5$ and the total number of possible outcomes is $n(S) = 6$.
$P(E) = \frac{n(E)}{n(S)} = \frac{5}{6}$.
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View Solution279
EasyMCQ
$A$ card is selected from a pack of $52$ cards. How many points are there in the sample space?
A
$52$
B
$26$
C
$13$
D
$4$
Solution
(A) standard pack of cards contains $52$ distinct cards.
When one card is selected from the pack,any of the $52$ cards can be chosen.
Therefore,the total number of possible outcomes in the sample space is $52$.
Thus,there are $52$ points in the sample space.
When one card is selected from the pack,any of the $52$ cards can be chosen.
Therefore,the total number of possible outcomes in the sample space is $52$.
Thus,there are $52$ points in the sample space.
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EasyMCQ
$A$ card is selected from a pack of $52$ cards. Calculate the probability that the card is an ace of spades.
A
$1/52$
B
$1/26$
C
$1/13$
D
$4/52$
Solution
(A) Let $A$ be the event in which the card drawn is an ace of spades.
In a standard pack of $52$ cards,there is only $1$ ace of spades.
Therefore,the number of favourable outcomes $n(A) = 1$.
The total number of possible outcomes $n(S) = 52$.
The probability $P(A)$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{1}{52}$.
In a standard pack of $52$ cards,there is only $1$ ace of spades.
Therefore,the number of favourable outcomes $n(A) = 1$.
The total number of possible outcomes $n(S) = 52$.
The probability $P(A)$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{1}{52}$.
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View Solution281
EasyMCQ
$A$ card is selected from a pack of $52$ cards. Calculate the probability that the card is an ace.
A
$1/13$
B
$1/26$
C
$4/13$
D
$1/52$
Solution
(A) Let $E$ be the event that the card drawn is an ace.
Since there are $4$ aces in a pack of $52$ cards,the number of favourable outcomes is $n(E) = 4$.
The total number of possible outcomes is $n(S) = 52$.
Therefore,the probability $P(E)$ is given by:
$P(E) = \frac{n(E)}{n(S)} = \frac{4}{52} = \frac{1}{13}$.
Since there are $4$ aces in a pack of $52$ cards,the number of favourable outcomes is $n(E) = 4$.
The total number of possible outcomes is $n(S) = 52$.
Therefore,the probability $P(E)$ is given by:
$P(E) = \frac{n(E)}{n(S)} = \frac{4}{52} = \frac{1}{13}$.
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EasyMCQ
$A$ card is selected from a pack of $52$ cards. Calculate the probability that the card is a black card.
A
$1/2$
B
$1/4$
C
$1/13$
D
$3/4$
Solution
(A) Let $F$ be the event in which the card drawn is black.
Since there are $26$ black cards in a pack of $52$ cards,$n(F) = 26$.
The total number of possible outcomes is $n(S) = 52$.
Therefore,the probability $P(F) = \frac{n(F)}{n(S)} = \frac{26}{52} = \frac{1}{2}$.
Since there are $26$ black cards in a pack of $52$ cards,$n(F) = 26$.
The total number of possible outcomes is $n(S) = 52$.
Therefore,the probability $P(F) = \frac{n(F)}{n(S)} = \frac{26}{52} = \frac{1}{2}$.
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EasyMCQ
$A$ fair coin with $1$ marked on one face and $6$ on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is $3$.
A
$\frac{1}{12}$
B
$\frac{1}{6}$
C
$\frac{1}{4}$
D
$\frac{1}{3}$
Solution
(A) The fair coin has faces marked $1$ and $6$. The fair die has faces marked $1, 2, 3, 4, 5,$ and $6$.
The sample space $S$ is the set of all possible ordered pairs $(c, d)$,where $c$ is the coin result and $d$ is the die result:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
The total number of outcomes is $n(S) = 12$.
Let $A$ be the event that the sum of the numbers is $3$. Looking at the sample space,the only outcome where the sum is $3$ is $(1, 2)$.
Thus,$A = \{(1, 2)\}$ and $n(A) = 1$.
The probability $P(A)$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{1}{12}$
The sample space $S$ is the set of all possible ordered pairs $(c, d)$,where $c$ is the coin result and $d$ is the die result:
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$
The total number of outcomes is $n(S) = 12$.
Let $A$ be the event that the sum of the numbers is $3$. Looking at the sample space,the only outcome where the sum is $3$ is $(1, 2)$.
Thus,$A = \{(1, 2)\}$ and $n(A) = 1$.
The probability $P(A)$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{1}{12}$
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EasyMCQ
$A$ fair coin with $1$ marked on one face and $6$ on the other and a fair die are both tossed. Find the probability that the sum of the numbers that turn up is $12$.
A
$\frac{1}{12}$
B
$\frac{1}{6}$
C
$\frac{1}{3}$
D
$\frac{1}{2}$
Solution
(A) The fair coin has faces marked $1$ and $6$. The fair die has faces marked $1, 2, 3, 4, 5,$ and $6$.
When both are tossed,the sample space $S$ consists of all possible pairs $(c, d)$ where $c \in \{1, 6\}$ and $d \in \{1, 2, 3, 4, 5, 6\}$.
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$.
The total number of outcomes is $n(S) = 12$.
Let $B$ be the event that the sum of the numbers is $12$.
The only outcome that satisfies this is $(6, 6)$,so $B = \{(6, 6)\}$.
The number of favorable outcomes is $n(B) = 1$.
Therefore,the probability $P(B) = \frac{n(B)}{n(S)} = \frac{1}{12}$.
When both are tossed,the sample space $S$ consists of all possible pairs $(c, d)$ where $c \in \{1, 6\}$ and $d \in \{1, 2, 3, 4, 5, 6\}$.
$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$.
The total number of outcomes is $n(S) = 12$.
Let $B$ be the event that the sum of the numbers is $12$.
The only outcome that satisfies this is $(6, 6)$,so $B = \{(6, 6)\}$.
The number of favorable outcomes is $n(B) = 1$.
Therefore,the probability $P(B) = \frac{n(B)}{n(S)} = \frac{1}{12}$.
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EasyMCQ
There are four men and six women on the city council. If one council member is selected for a committee at random,what is the probability that the selected member is a woman (in $/5$)?
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(C) The total number of council members is $4 + 6 = 10$.
Let $S$ be the sample space,so $n(S) = 10$.
Let $A$ be the event that the selected council member is a woman.
The number of women on the council is $6$,so $n(A) = 6$.
The probability $P(A)$ is given by the formula $P(A) = \frac{n(A)}{n(S)}$.
Therefore,$P(A) = \frac{6}{10} = \frac{3}{5}$.
Let $S$ be the sample space,so $n(S) = 10$.
Let $A$ be the event that the selected council member is a woman.
The number of women on the council is $6$,so $n(A) = 6$.
The probability $P(A)$ is given by the formula $P(A) = \frac{n(A)}{n(S)}$.
Therefore,$P(A) = \frac{6}{10} = \frac{3}{5}$.
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EasyMCQ
Three coins are tossed once. Find the probability of getting $3$ heads.
A
$1/8$
B
$3/8$
C
$1/2$
D
$7/8$
Solution
(A) When three coins are tossed once,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $E$ be the event of getting $3$ heads. The favorable outcome is:
$E = \{HHH\}$
Therefore,the number of favorable outcomes is $n(E) = 1$.
The probability of the event $E$ is given by:
$P(E) = \frac{n(E)}{n(S)} = \frac{1}{8}$
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $E$ be the event of getting $3$ heads. The favorable outcome is:
$E = \{HHH\}$
Therefore,the number of favorable outcomes is $n(E) = 1$.
The probability of the event $E$ is given by:
$P(E) = \frac{n(E)}{n(S)} = \frac{1}{8}$
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EasyMCQ
Three coins are tossed once. Find the probability of getting $2$ heads.
A
$3/8$
B
$1/8$
C
$1/2$
D
$1/4$
Solution
(A) When three coins are tossed once,the sample space is given by $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
$\therefore n(S) = 8$.
The probability of an event $A$ is given by $P(A) = \frac{n(A)}{n(S)}$.
Let $C$ be the event of getting exactly $2$ heads.
Then,$C = \{HHT, HTH, THH\}$.
$\therefore n(C) = 3$.
Thus,$P(C) = \frac{n(C)}{n(S)} = \frac{3}{8}$.
$\therefore n(S) = 8$.
The probability of an event $A$ is given by $P(A) = \frac{n(A)}{n(S)}$.
Let $C$ be the event of getting exactly $2$ heads.
Then,$C = \{HHT, HTH, THH\}$.
$\therefore n(C) = 3$.
Thus,$P(C) = \frac{n(C)}{n(S)} = \frac{3}{8}$.
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EasyMCQ
Three coins are tossed once. Find the probability of getting at least $2$ heads.
A
$\frac{1}{8}$
B
$\frac{1}{4}$
C
$\frac{1}{2}$
D
$\frac{3}{8}$
Solution
(C) When three coins are tossed once,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $D$ be the event of getting at least $2$ heads. The favorable outcomes are:
$D = \{HHH, HHT, HTH, THH\}$
Therefore,the number of favorable outcomes is $n(D) = 4$.
The probability of event $D$ is given by:
$P(D) = \frac{n(D)}{n(S)} = \frac{4}{8} = \frac{1}{2}$
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $D$ be the event of getting at least $2$ heads. The favorable outcomes are:
$D = \{HHH, HHT, HTH, THH\}$
Therefore,the number of favorable outcomes is $n(D) = 4$.
The probability of event $D$ is given by:
$P(D) = \frac{n(D)}{n(S)} = \frac{4}{8} = \frac{1}{2}$
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View Solution289
EasyMCQ
Three coins are tossed once. Find the probability of getting at most $2$ heads.
A
$1/8$
B
$3/8$
C
$7/8$
D
$1/2$
Solution
(C) When three coins are tossed once,the sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Therefore,the total number of outcomes is $n(S) = 8$.
Let $E$ be the event of getting at most $2$ heads. This means we exclude the case of $3$ heads $(HHH)$.
The favorable outcomes are $E = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Thus,$n(E) = 7$.
The probability is $P(E) = \frac{n(E)}{n(S)} = \frac{7}{8}$.
Therefore,the total number of outcomes is $n(S) = 8$.
Let $E$ be the event of getting at most $2$ heads. This means we exclude the case of $3$ heads $(HHH)$.
The favorable outcomes are $E = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Thus,$n(E) = 7$.
The probability is $P(E) = \frac{n(E)}{n(S)} = \frac{7}{8}$.
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View Solution290
EasyMCQ
Three coins are tossed once. Find the probability of getting no head.
A
$1/8$
B
$1/4$
C
$3/8$
D
$1/2$
Solution
(A) When three coins are tossed once,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $F$ be the event of getting no head. The only outcome corresponding to this event is:
$F = \{TTT\}$
Thus,the number of favourable outcomes is $n(F) = 1$.
The probability of event $F$ is given by:
$P(F) = \frac{n(F)}{n(S)} = \frac{1}{8}$
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $F$ be the event of getting no head. The only outcome corresponding to this event is:
$F = \{TTT\}$
Thus,the number of favourable outcomes is $n(F) = 1$.
The probability of event $F$ is given by:
$P(F) = \frac{n(F)}{n(S)} = \frac{1}{8}$
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View Solution291
EasyMCQ
Three coins are tossed once. Find the probability of getting $3$ tails.
A
$1/8$
B
$3/8$
C
$1/2$
D
$7/8$
Solution
(A) When three coins are tossed once,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $G$ be the event of getting $3$ tails. The favorable outcome is:
$G = \{TTT\}$
Therefore,the number of favorable outcomes is $n(G) = 1$.
The probability of event $G$ is given by:
$P(G) = \frac{n(G)}{n(S)} = \frac{1}{8}$
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $G$ be the event of getting $3$ tails. The favorable outcome is:
$G = \{TTT\}$
Therefore,the number of favorable outcomes is $n(G) = 1$.
The probability of event $G$ is given by:
$P(G) = \frac{n(G)}{n(S)} = \frac{1}{8}$
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View Solution292
EasyMCQ
Three coins are tossed once. Find the probability of getting exactly $2$ tails.
A
$1/8$
B
$3/8$
C
$1/2$
D
$3/4$
Solution
(B) When three coins are tossed once,the sample space is given by $S = \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}$.
$\therefore n(S) = 8$.
The probability of an event $A$ is given by $P(A) = \frac{n(A)}{n(S)}$.
Let $E$ be the event of getting exactly $2$ tails.
The favorable outcomes are $E = \{ HTT, THT, TTH \}$.
$\therefore n(E) = 3$.
Thus,$P(E) = \frac{n(E)}{n(S)} = \frac{3}{8}$.
$\therefore n(S) = 8$.
The probability of an event $A$ is given by $P(A) = \frac{n(A)}{n(S)}$.
Let $E$ be the event of getting exactly $2$ tails.
The favorable outcomes are $E = \{ HTT, THT, TTH \}$.
$\therefore n(E) = 3$.
Thus,$P(E) = \frac{n(E)}{n(S)} = \frac{3}{8}$.
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View Solution293
EasyMCQ
Three coins are tossed once. Find the probability of getting no tails.
A
$1/8$
B
$3/8$
C
$1/2$
D
$7/8$
Solution
(A) When three coins are tossed once,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $A$ be the event of getting no tails. This means all three coins must show heads.
$A = \{HHH\}$
Therefore,the number of favorable outcomes is $n(A) = 1$.
The probability of event $A$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{1}{8}$
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Therefore,the total number of possible outcomes is $n(S) = 8$.
Let $A$ be the event of getting no tails. This means all three coins must show heads.
$A = \{HHH\}$
Therefore,the number of favorable outcomes is $n(A) = 1$.
The probability of event $A$ is given by:
$P(A) = \frac{n(A)}{n(S)} = \frac{1}{8}$
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View Solution294
EasyMCQ
Three coins are tossed once. Find the probability of getting at most two tails.
A
$1/8$
B
$3/8$
C
$7/8$
D
$1/2$
Solution
(C) When three coins are tossed once,the sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Therefore,the total number of outcomes is $n(S) = 8$.
Let $J$ be the event of getting at most $2$ tails. This means we exclude the case of $3$ tails $(TTT)$.
Thus,$J = \{HHH, HHT, HTH, THH, HTT, THT, TTH\}$.
Therefore,$n(J) = 7$.
The probability of event $J$ is $P(J) = \frac{n(J)}{n(S)} = \frac{7}{8}$.
Therefore,the total number of outcomes is $n(S) = 8$.
Let $J$ be the event of getting at most $2$ tails. This means we exclude the case of $3$ tails $(TTT)$.
Thus,$J = \{HHH, HHT, HTH, THH, HTT, THT, TTH\}$.
Therefore,$n(J) = 7$.
The probability of event $J$ is $P(J) = \frac{n(J)}{n(S)} = \frac{7}{8}$.
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View Solution295
EasyMCQ
$A$ letter is chosen at random from the word $ASSASSINATION$. Find the probability that the letter is a vowel. (in $/13$)
A
$6$
B
$5$
C
$7$
D
$8$
Solution
(A) The word $ASSASSINATION$ contains $13$ letters in total.
Therefore,the total number of outcomes $n(S) = 13$.
The vowels in the word $ASSASSINATION$ are $A, A, I, A, I, O$.
There are $6$ vowels in the word.
Therefore,the number of favorable outcomes $n(E) = 6$.
The probability of choosing a vowel is $P(E) = \frac{n(E)}{n(S)} = \frac{6}{13}$.
Therefore,the total number of outcomes $n(S) = 13$.
The vowels in the word $ASSASSINATION$ are $A, A, I, A, I, O$.
There are $6$ vowels in the word.
Therefore,the number of favorable outcomes $n(E) = 6$.
The probability of choosing a vowel is $P(E) = \frac{n(E)}{n(S)} = \frac{6}{13}$.
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View Solution296
EasyMCQ
$A$ letter is chosen at random from the word $ASSASSINATION$. Find the probability that the letter is a consonant. (in $/13$)
A
$6$
B
$7$
C
$8$
D
$9$
Solution
(B) The word $ASSASSINATION$ contains $13$ letters in total.
The letters are $A, S, S, A, S, S, I, N, A, T, I, O, N$.
The vowels are $A, A, I, A, I, O$ (total $6$ vowels).
The consonants are $S, S, S, S, N, T, N$ (total $7$ consonants).
Therefore,the probability of choosing a consonant is $\frac{\text{Number of consonants}}{\text{Total number of letters}} = \frac{7}{13}$.
The letters are $A, S, S, A, S, S, I, N, A, T, I, O, N$.
The vowels are $A, A, I, A, I, O$ (total $6$ vowels).
The consonants are $S, S, S, S, N, T, N$ (total $7$ consonants).
Therefore,the probability of choosing a consonant is $\frac{\text{Number of consonants}}{\text{Total number of letters}} = \frac{7}{13}$.
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View Solution297
Easy
Check whether the following probabilities $P(A)$ and $P(B)$ are consistently defined: $P(A) = 0.5$,$P(B) = 0.7$,$P(A \cap B) = 0.6$.
Solution
(N/A) Given: $P(A) = 0.5$,$P(B) = 0.7$,and $P(A \cap B) = 0.6$.
It is a fundamental property of probability that for any two events $A$ and $B$,the intersection of the events must be a subset of each individual event,i.e.,$(A \cap B) \subseteq A$ and $(A \cap B) \subseteq B$.
Consequently,the probability of the intersection must satisfy $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$.
In this case,we observe that $P(A \cap B) = 0.6$ and $P(A) = 0.5$.
Since $0.6 > 0.5$,the condition $P(A \cap B) \leq P(A)$ is violated.
Therefore,the given probabilities $P(A)$ and $P(B)$ are not consistently defined.
It is a fundamental property of probability that for any two events $A$ and $B$,the intersection of the events must be a subset of each individual event,i.e.,$(A \cap B) \subseteq A$ and $(A \cap B) \subseteq B$.
Consequently,the probability of the intersection must satisfy $P(A \cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$.
In this case,we observe that $P(A \cap B) = 0.6$ and $P(A) = 0.5$.
Since $0.6 > 0.5$,the condition $P(A \cap B) \leq P(A)$ is violated.
Therefore,the given probabilities $P(A)$ and $P(B)$ are not consistently defined.
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Easy
Check whether the following probabilities $P(A) = 0.5$,$P(B) = 0.4$,and $P(A \cup B) = 0.8$ are consistently defined.
Solution
(A) Given: $P(A) = 0.5$,$P(B) = 0.4$,and $P(A \cup B) = 0.8$.
For any two events $A$ and $B$,the probability of their union is given by the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $0.8 = 0.5 + 0.4 - P(A \cap B)$.
$0.8 = 0.9 - P(A \cap B) \implies P(A \cap B) = 0.9 - 0.8 = 0.1$.
Since $0 \leq P(A \cap B) \leq P(A)$ and $0 \leq P(A \cap B) \leq P(B)$,and here $0.1 \leq 0.5$ and $0.1 \leq 0.4$,the values are consistent.
For any two events $A$ and $B$,the probability of their union is given by the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $0.8 = 0.5 + 0.4 - P(A \cap B)$.
$0.8 = 0.9 - P(A \cap B) \implies P(A \cap B) = 0.9 - 0.8 = 0.1$.
Since $0 \leq P(A \cap B) \leq P(A)$ and $0 \leq P(A \cap B) \leq P(B)$,and here $0.1 \leq 0.5$ and $0.1 \leq 0.4$,the values are consistent.
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Easy
Events $E$ and $F$ are such that $P(\text{not } E \text{ and not } F) = 0.25$. State whether $E$ and $F$ are mutually exclusive.
Solution
(N/A) It is given that $P(\text{not } E \text{ or not } F) = 0.25$.
By De Morgan's Law,this is equivalent to $P((E \cap F)') = 0.25$.
We know that $P(E \cap F) = 1 - P((E \cap F)')$.
Substituting the value,we get $P(E \cap F) = 1 - 0.25 = 0.75$.
Since $P(E \cap F) = 0.75 \neq 0$,the events $E$ and $F$ are not mutually exclusive.
By De Morgan's Law,this is equivalent to $P((E \cap F)') = 0.25$.
We know that $P(E \cap F) = 1 - P((E \cap F)')$.
Substituting the value,we get $P(E \cap F) = 1 - 0.25 = 0.75$.
Since $P(E \cap F) = 0.75 \neq 0$,the events $E$ and $F$ are not mutually exclusive.
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View Solution300
MediumMCQ
The probability that a student will pass the final examination in both English and Hindi is $0.5$ and the probability of passing neither is $0.1$. If the probability of passing the English examination is $0.75$,what is the probability of passing the Hindi examination?
A
$0.65$
B
$0.55$
C
$0.45$
D
$0.35$
Solution
(A) Let $A$ and $B$ be the events of passing English and Hindi examinations respectively.
Given: $P(A \cap B) = 0.5$ and $P(A' \cap B') = 0.1$.
We know that $P(A \cup B)' = P(A' \cap B') = 0.1$ by De Morgan's law.
Therefore,$P(A \cup B) = 1 - P(A \cup B)' = 1 - 0.1 = 0.9$.
We also know the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given $P(A) = 0.75$,we substitute the values:
$0.9 = 0.75 + P(B) - 0.5$.
$0.9 = 0.25 + P(B)$.
$P(B) = 0.9 - 0.25 = 0.65$.
Thus,the probability of passing the Hindi examination is $0.65$.
Given: $P(A \cap B) = 0.5$ and $P(A' \cap B') = 0.1$.
We know that $P(A \cup B)' = P(A' \cap B') = 0.1$ by De Morgan's law.
Therefore,$P(A \cup B) = 1 - P(A \cup B)' = 1 - 0.1 = 0.9$.
We also know the formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Given $P(A) = 0.75$,we substitute the values:
$0.9 = 0.75 + P(B) - 0.5$.
$0.9 = 0.25 + P(B)$.
$P(B) = 0.9 - 0.25 = 0.65$.
Thus,the probability of passing the Hindi examination is $0.65$.
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View SolutionProbability — Set Based probability · Frequently Asked Questions
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