Mix Examples - Probability Questions in English

(C) standard deck contains $52$ cards.
Picture cards (also known as face cards) are the King,Queen,and Jack of each of the $4$ suits.
Total number of picture cards = $3 \times 4 = 12$.
The probability $P$ of drawing a picture card is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$P = \frac{12}{52}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $P = \frac{3}{13}$.

(D) In a leap year,the month of February has $29$ days.
$29$ days consist of $4$ complete weeks and $1$ extra day.
$4$ weeks contain $4$ Mondays.
The $29^{th}$ day can be any one of the $7$ days of the week (Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,or Sunday).
For February to have $5$ Mondays,the $29^{th}$ day must be a Monday.
Since there is only $1$ favorable outcome (Monday) out of $7$ possible outcomes,the probability is $\frac{1}{7}$.

(A) Total number of students = $50$.
Number of girls = $22$.
Number of boys = $\text{Total students} - \text{Number of girls} = 50 - 22 = 28$.
The probability that a student who gets the first rank is a boy is given by the ratio of the number of boys to the total number of students.
Probability = $\frac{\text{Number of boys}}{\text{Total number of students}} = \frac{28}{50}$.
To convert this to a decimal, multiply the numerator and denominator by $2$: $\frac{28 \times 2}{50 \times 2} = \frac{56}{100} = 0.56$.

(B) standard pack of cards contains $52$ cards in total.
There are two red suits: Hearts and Diamonds.
Each suit contains $13$ cards.
Therefore,the total number of red cards is $13 + 13 = 26$.
The probability of drawing a red card is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Probability $= \frac{\text{Number of red cards}}{\text{Total number of cards}} = \frac{26}{52} = \frac{1}{2}$.

(C) non-leap year has $365$ days.
$365$ days $= 52$ weeks and $1$ extra day.
$52$ weeks contain $52$ Sundays.
The remaining $1$ day can be any of the $7$ days of the week: {Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday}.
For the year to have $53$ Sundays,the extra day must be a Sunday.
There is only $1$ favorable outcome out of $7$ possible outcomes.
Therefore,the probability is $\frac{1}{7}$.

(D) The month of August has $31$ days.
$31$ days consist of $4$ full weeks and $3$ extra days $(31 = 4 \times 7 + 3)$.
The $3$ extra days can be any of the following sets of consecutive days:
$1$. (Sunday,Monday,Tuesday)
$2$. (Monday,Tuesday,Wednesday)
$3$. (Tuesday,Wednesday,Thursday)
$4$. (Wednesday,Thursday,Friday)
$5$. (Thursday,Friday,Saturday)
$6$. (Friday,Saturday,Sunday)
$7$. (Saturday,Sunday,Monday)
Out of these $7$ possibilities,the sets containing Wednesday are: (Monday,Tuesday,Wednesday),(Tuesday,Wednesday,Thursday),and (Wednesday,Thursday,Friday).
There are $3$ favorable outcomes out of $7$ total possibilities.
Therefore,the probability is $\frac{3}{7}$.

(A) Total number of days = $30$.
Number of days the forecast was true = $20$.
Number of days the forecast was false = $\text{Total days} - \text{True days} = 30 - 20 = 10$.
The probability that the forecast proved to be false is given by the ratio of the number of days the forecast was false to the total number of days.
$P(\text{False}) = \frac{10}{30} = \frac{1}{3}$.

(B) leap year has $366$ days.
$366$ days $= 52$ weeks and $2$ extra days.
These $2$ extra days can be any of the following pairs: (Sunday,Monday),(Monday,Tuesday),(Tuesday,Wednesday),(Wednesday,Thursday),(Thursday,Friday),(Friday,Saturday),or (Saturday,Sunday).
There are $7$ possible outcomes for these $2$ extra days.
For the year to have $53$ Sundays,one of the $2$ extra days must be a Sunday.
The favorable outcomes are (Sunday,Monday) and (Saturday,Sunday).
There are $2$ favorable outcomes.
Therefore,the probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{7}$.

(C) certain event is an event that is sure to occur.
By definition,the probability of an event that is certain to happen is $1$.
Therefore,the correct option is $C$.

(D) An impossible event is an event that cannot occur.
By definition,the probability of an event $E$ lies in the range $0 \le P(E) \le 1$.
Since an impossible event has no chance of occurring,its probability is defined as $0$.

(A) The probability of an event $A$ occurring is denoted by $P(A) = 0.43$.
The probability of the event $A$ not occurring (complementary event) is denoted by $P(\text{not } A)$ or $P(A')$.
The sum of the probabilities of an event occurring and not occurring is always $1$.
$P(A) + P(A') = 1$
$0.43 + P(A') = 1$
$P(A') = 1 - 0.43$
$P(A') = 0.57$
Therefore,the probability of event $A$ not occurring is $0.57$.

(B) The probability of any event $E$,denoted by $P(E)$,always lies in the range $0 \le P(E) \le 1$.
Therefore,the minimum possible value of the probability of any event is $0$ (which corresponds to an impossible event),and the maximum possible value is $1$ (which corresponds to a sure event).
Thus,the correct option is $B$.

(C) The probability of any event $E$,denoted by $P(E)$,always lies in the range $0 \le P(E) \le 1$.
This means the minimum possible value of a probability is $0$ (for an impossible event) and the maximum possible value is $1$ (for a sure event).
Therefore,the maximum value of the probability of any event is $1$.

(D) In a test of $100$ marks,the possible marks a student can score range from $0$ to $100$.
Total number of possible outcomes = $100 - 0 + 1 = 101$.
Since each score is considered an equally likely outcome,the probability of scoring exactly $70$ marks is the ratio of the number of favorable outcomes to the total number of possible outcomes.
Number of favorable outcomes (scoring $70$) = $1$.
Therefore,the probability $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{101}$.

(A) The single-digit natural numbers are the set $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
The total number of outcomes is $n(S) = 9$.
The multiples of $3$ in this set are $E = \{3, 6, 9\}$.
The number of favorable outcomes is $n(E) = 3$.
The probability $P(E)$ is given by the ratio of favorable outcomes to total outcomes:
$P(E) = \frac{n(E)}{n(S)} = \frac{3}{9} = \frac{1}{3}$.

(B) balanced die has $6$ faces,numbered $1, 2, 3, 4, 5,$ and $6$.
Total number of possible outcomes $n(S) = 6$.
The event of getting $3$ is $E = {3}$.
Number of favorable outcomes $n(E) = 1$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Therefore,$P(E) = \frac{1}{6}$.

(C) The event 'the sun rises in the east' is a certain event or a sure event.
In probability theory,the probability of a certain event is always $1$.

(D) An acacia tree is a distinct species from a mango tree. It is biologically impossible for a mango to grow on an acacia tree. Since this is an impossible event,its probability is $0$.

(A) The total outcomes when a balanced die is thrown are $S = \{1, 2, 3, 4, 5, 6\}$.
The total number of outcomes is $n(S) = 6$.
$A$ composite number is a positive integer greater than $1$ that has at least one divisor other than $1$ and itself.
In the set $\{1, 2, 3, 4, 5, 6\}$,the composite numbers are $4$ and $6$.
Thus,the favorable outcomes are $E = \{4, 6\}$.
The number of favorable outcomes is $n(E) = 2$.
The probability $P(E)$ is given by $\frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}$.

(B) The total number of cards in a well-shuffled pack is $n(S) = 52$.
There are $4$ queens in a standard deck of $52$ cards (one for each suit: hearts,diamonds,clubs,and spades).
Let $E$ be the event of drawing a queen. The number of favorable outcomes is $n(E) = 4$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Substituting the values,we get $P(E) = \frac{4}{52} = \frac{1}{13}$.

(C) The total number of possible outcomes when a balanced die is thrown is $S = \{1, 2, 3, 4, 5, 6\}$.
Thus,the total number of outcomes is $n(S) = 6$.
The odd numbers on a die are $1, 3,$ and $5$.
Let $E$ be the event of getting an odd number,so $E = \{1, 3, 5\}$.
The number of favorable outcomes is $n(E) = 3$.
The probability of getting an odd number is $P(E) = \frac{n(E)}{n(S)} = \frac{3}{6} = \frac{1}{2}$.

(D) Total number of students = $30$.
Number of boys = $12$.
Number of girls = $30 - 12 = 18$.
The probability of winning the first prize by a girl is the ratio of the number of girls to the total number of students.
Probability = $\frac{\text{Number of girls}}{\text{Total students}} = \frac{18}{30}$.
Simplifying the fraction by dividing both numerator and denominator by $6$,we get $\frac{18 \div 6}{30 \div 6} = \frac{3}{5}$.
Therefore,the correct option is $D$.

(A) When two coins are tossed simultaneously,the sample space $S$ is given by: $S = \{(H, H), (H, T), (T, H), (T, T)\}$.
The total number of possible outcomes is $n(S) = 4$.
We want to find the probability of getting no tail. The outcome with no tail is $(H, H)$.
Thus,the number of favorable outcomes is $n(E) = 1$.
The probability $P(E)$ is given by the formula: $P(E) = \frac{n(E)}{n(S)} = \frac{1}{4}$.

(B) In any given year,$26$ January can fall on any of the $7$ days of the week (Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,or Sunday) with equal probability.
Since there are $7$ possible outcomes and only $1$ of them is Sunday,the probability is calculated as:
$P(\text{Sunday}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{7}$.

(C) The total number of women is $700$.
The number of employed women is $280$.
The probability $P$ of selecting an employed woman is given by the ratio of the number of employed women to the total number of women.
$P = \frac{\text{Number of employed women}}{\text{Total number of women}}$
$P = \frac{280}{700}$
Dividing both numerator and denominator by $140$,we get:
$P = \frac{2}{5}$

(D) Total number of students = $40$.
Number of girls = $16$.
Number of boys = $40 - 16 = 24$.
The probability of selecting a boy is given by the ratio of the number of boys to the total number of students.
$P(\text{boy}) = \frac{\text{Number of boys}}{\text{Total number of students}} = \frac{24}{40}$.
Dividing both numerator and denominator by $8$, we get $\frac{24}{40} = \frac{3}{5}$.
Therefore, the probability that a boy is selected is $3/5$.

(A) Total number of tosses = $200$.
Number of times tail occurs = $92$.
Number of times head occurs = $200 - 92 = 108$.
The probability of getting a head is given by the ratio of the number of heads to the total number of tosses.
$P(\text{Head}) = \frac{\text{Number of heads}}{\text{Total number of tosses}} = \frac{108}{200}$.
$P(\text{Head}) = \frac{54}{100} = 0.54$.

(B) The total number of marks available in the examination is $50$.
It is impossible to obtain $51$ marks out of $50$ marks because the maximum possible score is $50$.
Since the event of getting $51$ marks is an impossible event,its probability is $0$.