Mix Examples - Probability Questions in English

(A) When two balanced dice are rolled,the total number of possible outcomes is $6 \times 6 = 36$.
The outcomes where the numbers on both dice are equal are: $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$.
There are $6$ such favorable outcomes.
The probability $P$ of getting equal numbers is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{6}{36} = \frac{1}{6}$.

(B) Total number of screws = $500$.
Number of defective screws = $80$.
Number of non-defective screws = $500 - 80 = 420$.
The probability of selecting a non-defective screw is given by the ratio of the number of non-defective screws to the total number of screws.
Probability = $\frac{\text{Number of non-defective screws}}{\text{Total number of screws}} = \frac{420}{500}$.
Simplifying the fraction: $\frac{420}{500} = \frac{42}{50} = \frac{21}{25}$.
Therefore,the probability is $\frac{21}{25}$.

(C) The total number of chits in the box is $n(S) = 20$.
The chits are numbered from $1$ to $20$.
The odd numbers between $1$ and $20$ are $1, 3, 5, 7, 9, 11, 13, 15, 17, 19$.
The number of odd-numbered chits is $n(E) = 10$.
The probability $P(E)$ of drawing an odd-numbered chit is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Substituting the values,we get $P(E) = \frac{10}{20} = \frac{1}{2}$.

(D) The total number of outcomes is $20$ (since chits are numbered $1$ to $20$).
Prime numbers between $1$ and $20$ are: $2, 3, 5, 7, 11, 13, 17, 19$.
The number of favorable outcomes is $8$.
The probability $P$ of drawing a prime number is given by the ratio of favorable outcomes to the total number of outcomes.
$P = \frac{8}{20} = \frac{2}{5}$.

(A) Total number of chits in the box = $20$.
So,the total number of possible outcomes = $20$.
The multiples of $5$ between $1$ and $20$ are $5, 10, 15, 20$.
Number of favorable outcomes = $4$.
The probability of an event is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
Therefore,$P(E) = \frac{4}{20} = \frac{1}{5}$.

(B) When an unbiased coin is tossed $3$ times,the total number of possible outcomes is $2^3 = 8$.
The sample space $S$ is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
The event $A$ is defined as getting exactly $2$ heads.
The favorable outcomes for event $A$ are $\{HHT, HTH, THH\}$.
The number of favorable outcomes $n(A) = 3$.
Therefore,the probability $P(A) = \frac{n(A)}{n(S)} = \frac{3}{8}$.

(C) When an unbiased coin is tossed thrice,the total number of possible outcomes is $2^3 = 8$.
The sample space $S$ is given by: $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$.
Event $B$ is defined as getting at least $2$ heads.
The favorable outcomes for event $B$ are: $\{HHH, HHT, HTH, THH\}$.
The number of favorable outcomes is $n(B) = 4$.
The probability of event $B$ is $P(B) = \frac{n(B)}{n(S)} = \frac{4}{8} = \frac{1}{2}$.

(D) When an unbiased coin is tossed thrice,the sample space $S$ is given by:
$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$
Total number of outcomes $n(S) = 8$.
Event $C$ is defined as getting at most $2$ heads.
This means we can have $0, 1,$ or $2$ heads.
The outcomes corresponding to event $C$ are:
$0$ head: ${TTT}$
$1$ head: ${HTT, THT, TTH}$
$2$ heads: ${HHT, HTH, THH}$
Total favorable outcomes $n(C) = 1 + 3 + 3 = 7$.
Alternatively,the only outcome not included is ${HHH}$ (which has $3$ heads).
Therefore,$P(C) = \frac{n(C)}{n(S)} = \frac{7}{8}$.

(A) Total number of boards in the box = $50$.
Thus,the total number of possible outcomes = $50$.
Let $E$ be the event of selecting a board with a number that is a multiple of $11$.
The multiples of $11$ between $1$ and $50$ are $11, 22, 33,$ and $44$.
Number of favorable outcomes = $4$.
The probability $P(E)$ is given by the formula: $P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
$P(E) = \frac{4}{50} = \frac{2}{25}$.

(B) The total number of outcomes is $50$ (since boards are numbered $1$ to $50$).
Prime numbers between $1$ and $50$ are: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$.
Counting these,we find there are $15$ prime numbers.
The probability $P$ of selecting a prime number is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{15}{50} = \frac{3}{10}$.

(C) Total number of boards = $50$.
The multiples of $10$ between $1$ and $50$ are: $10, 20, 30, 40, 50$.
Number of favorable outcomes = $5$.
Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{50} = \frac{1}{10}$.

(D) Total number of balls in the box = $5$ (white) + $7$ (red) + $4$ (black) + $2$ (blue) = $18$.
The number of favorable outcomes (selecting either a black or white ball) = $5$ (white) + $4$ (black) = $9$.
The probability $P$ of selecting a black or white ball is given by the ratio of favorable outcomes to the total number of outcomes.
$P = \frac{9}{18} = \frac{1}{2}$.

(A) The total number of balls in the box is the sum of all individual colored balls:
Total balls $= 5 + 7 + 4 + 2 = 18$.
The number of red balls is $7$.
The probability of selecting a red ball is given by the ratio of the number of red balls to the total number of balls:
$P(\text{Red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{7}{18}$.

(B) Total number of balls = $5 + 7 + 4 + 2 = 18$.
Number of blue balls = $2$.
Probability of selecting a blue ball $P(\text{Blue}) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{2}{18} = \frac{1}{9}$.
The probability that the ball is not blue is given by $P(\text{Not Blue}) = 1 - P(\text{Blue})$.
$P(\text{Not Blue}) = 1 - \frac{1}{9} = \frac{8}{9}$.

(C) $1$. First,calculate the area of the square board: $\text{Area of square} = \text{side}^2 = 20 \, cm \times 20 \, cm = 400 \, cm^2$.
$2$. Next,identify the type of triangle. Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,the triangle is a right-angled triangle.
$3$. Calculate the area of the triangular target: $\text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \, cm \times 12 \, cm = 30 \, cm^2$.
$4$. The probability $P$ that a dart hitting the board hits the target is given by the ratio of the area of the target to the area of the board: $P = \frac{\text{Area of triangle}}{\text{Area of square}} = \frac{30}{400} = \frac{3}{40}$.

(D) The set of one-digit natural numbers is $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
The total number of outcomes is $n(S) = 9$.
The set of even numbers among these is $E = \{2, 4, 6, 8\}$.
The number of favorable outcomes is $n(E) = 4$.
The probability $P(E)$ is given by $\frac{n(E)}{n(S)} = \frac{4}{9}$.

(A) The set of one-digit natural numbers is $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
The total number of outcomes is $n(S) = 9$.
The multiples of $3$ among these are $E = \{3, 6, 9\}$.
The number of favorable outcomes is $n(E) = 3$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Therefore,$P(E) = \frac{3}{9} = \frac{1}{3}$.

(B) The total number of possible outcomes when a balanced die is rolled is $S = \{1, 2, 3, 4, 5, 6\}$,so the total number of outcomes $n(S) = 6$.
The multiples of $3$ present on the faces of the die are $3$ and $6$.
Let $E$ be the event of getting a multiple of $3$. Then $E = \{3, 6\}$,so the number of favorable outcomes $n(E) = 2$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
$P(E) = \frac{2}{6} = \frac{1}{3}$.
Therefore,the probability of getting a multiple of $3$ is $\frac{1}{3}$.

(C) The total number of possible outcomes when a balanced die is rolled is $6$,which are $\{1, 2, 3, 4, 5, 6\}$.
Let $E$ be the event of getting a number greater than $4$.
The numbers greater than $4$ on a die are $5$ and $6$.
Thus,the number of favorable outcomes is $2$.
The probability $P(E)$ is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$P(E) = \frac{2}{6} = \frac{1}{3}$.

(D) non-leap year has $365$ days.
$365$ days $= 52$ weeks and $1$ extra day.
$52$ weeks contain $52$ Fridays.
For the year to have $53$ Fridays,the remaining $1$ extra day must be a Friday.
The possible outcomes for the extra day are: {Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday}.
Total number of outcomes $= 7$.
Number of favorable outcomes (Friday) $= 1$.
Therefore,the probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{7}$.

(A) The month of December has $31$ days.
$31$ days consist of $4$ full weeks and $3$ extra days $(31 = 7 \times 4 + 3)$.
The $3$ extra days can be any of the following combinations: (Sunday,Monday,Tuesday),(Monday,Tuesday,Wednesday),(Tuesday,Wednesday,Thursday),(Wednesday,Thursday,Friday),(Thursday,Friday,Saturday),(Friday,Saturday,Sunday),or (Saturday,Sunday,Monday).
Out of these $7$ possible outcomes,the combinations containing a Saturday are: (Thursday,Friday,Saturday),(Friday,Saturday,Sunday),and (Saturday,Sunday,Monday).
There are $3$ favorable outcomes out of $7$ total possibilities.
Therefore,the probability is $\frac{3}{7}$.

(B) non-leap year has $28$ days in the month of February.
$28$ days consist of exactly $4$ weeks ($4 \times 7 = 28$ days).
Therefore,every day of the week occurs exactly $4$ times in February of a non-leap year.
Since there are no extra days,it is impossible to have $5$ Wednesdays.
Thus,the probability is $0$.

(C) In a leap year,the month of February has $29$ days.
$29$ days consist of $4$ complete weeks and $1$ extra day $(29 = 4 \times 7 + 1)$.
These $4$ weeks contain $4$ Tuesdays,$4$ Wednesdays,$4$ Thursdays,$4$ Fridays,$4$ Saturdays,$4$ Sundays,and $4$ Mondays.
The remaining $1$ extra day can be any of the $7$ days of the week: {Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday}.
For February to have $5$ Tuesdays,this extra day must be a Tuesday.
There is only $1$ favorable outcome (Tuesday) out of $7$ possible outcomes.
Therefore,the probability is $\frac{1}{7}$.

(D) The sum of the probability of an event and its complement is always $1$.
Given $P(A) = 0.54$.
We know that $P(A) + P(\overline{A}) = 1$.
Therefore,$P(\overline{A}) = 1 - P(A)$.
$P(\overline{A}) = 1 - 0.54 = 0.46$.

(A) We know that for any event $B$,the sum of the probability of the event and its complement is $1$.
$P(B) + P(\overline{B}) = 1$
Given that $P(\overline{B}) = 0.81$.
Substituting the value,we get:
$P(B) + 0.81 = 1$
$P(B) = 1 - 0.81$
$P(B) = 0.19$

(B) The sum of the probability of an event and its complement is always $1$.
Given $P(C) = \frac{2}{5}$.
We know that $P(C) + P(\bar{C}) = 1$.
Therefore,$P(\bar{C}) = 1 - P(C)$.
$P(\bar{C}) = 1 - \frac{2}{5} = \frac{5 - 2}{5} = \frac{3}{5}$.

(C) The sum of the probability of an event and its complement is always $1$.
That is,$P(D) + P(\overline{D}) = 1$.
Given that $P(\overline{D}) = \frac{6}{17}$,we can substitute this value into the equation:
$P(D) + \frac{6}{17} = 1$
$P(D) = 1 - \frac{6}{17}$
$P(D) = \frac{17 - 6}{17}$
$P(D) = \frac{11}{17}$

(D) The month of April has $30$ days.
$30$ days consist of $4$ full weeks and $2$ extra days.
$4 \times 7 = 28$ days.
$30 - 28 = 2$ days remain.
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 month to have $5$ Fridays,the $2$ extra days must include a Friday.
The favorable outcomes are (Thursday,Friday) and (Friday,Saturday).
There are $2$ favorable outcomes.
Therefore,the probability is $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{7}$.

(A) The month of April has $30$ days.
$30$ days consist of $4$ complete weeks and $2$ extra days $(30 = 7 \times 4 + 2)$.
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 month to have $5$ 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 out of $7$ total possibilities.
Therefore,the probability is $2/7$.

(B) In a test of $100$ marks,the possible marks a student can score range from $0$ to $100$.
Total number of possible outcomes = $100 - 0 + 1 = 101$.
We need to find the probability of scoring marks more than $60$ and less than $66$.
The possible marks are $61, 62, 63, 64, 65$.
Number of favorable outcomes = $5$.
Probability = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{5}{101}$.

(C) The event of the Sun setting in the west is a 'sure event' or 'certain event'.
By definition,the probability of a sure event is always $1$.

(D) The total number of cards in a well-shuffled pack is $52$.
There are $4$ jacks in a standard deck of $52$ cards (one for each suit: hearts,diamonds,clubs,and spades).
The probability $P$ of an event is given by the formula: $P = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$.
Here,the number of favorable outcomes (drawing a jack) is $4$.
The total number of possible outcomes is $52$.
Therefore,the probability $P = \frac{4}{52} = \frac{1}{13}$.

(A) When a balanced die is rolled,the total number of possible outcomes is $S = \{1, 2, 3, 4, 5, 6\}$.
The total number of outcomes is $n(S) = 6$.
Let $E$ be the event of getting a number smaller than $3$.
The favorable outcomes are $E = \{1, 2\}$.
The number of favorable outcomes is $n(E) = 2$.
The probability $P(E)$ is given by the formula $P(E) = \frac{n(E)}{n(S)}$.
Therefore,$P(E) = \frac{2}{6} = \frac{1}{3}$.

(B) We know that the sum of the probability of an event and its complement is $1$,i.e.,$P(A) + P(\bar{A}) = 1$.
Given the ratio $P(A) : P(\bar{A}) = 3 : 2$,let $P(A) = 3x$ and $P(\bar{A}) = 2x$.
Substituting these into the equation: $3x + 2x = 1$.
$5x = 1$,which gives $x = \frac{1}{5}$.
Therefore,$P(A) = 3x = 3 \times \frac{1}{5} = \frac{3}{5}$.

(C) We know that for any event $A$,the sum of the probability of the event and its complement is $1$,i.e.,$P(A) + P(\overline{A}) = 1$.
Given the ratio $P(A) : P(\overline{A}) = 4 : 1$,we can let $P(A) = 4x$ and $P(\overline{A}) = 1x$ for some constant $x$.
Substituting these into the sum equation: $4x + 1x = 1$.
$5x = 1$,which gives $x = \frac{1}{5}$.
Therefore,$P(\overline{A}) = 1x = 1 \times \frac{1}{5} = \frac{1}{5}$.

(C) When an unbiased coin is tossed once,the possible outcomes are Head $(H)$ and Tail $(T)$.
When the same coin is tossed twice,the sample space $S$ is given by the set of all possible ordered pairs of outcomes.
$S = \{(H, H), (H, T), (T, H), (T, T)\}$.
Counting these,we find that there are $4$ possible outcomes.
Therefore,the total number of outcomes is $4$.

(B) When an unbiased coin is tossed twice,the sample space $S$ is given by: $S = \{HH, HT, TH, TT\}$.
The total number of possible outcomes is $n = 4$.
Let $A$ be the event of getting two heads. The only favourable outcome is $HH$.
The number of favourable outcomes is $m = 1$.
The probability of event $A$ is given by the formula $P(A) = \frac{m}{n}$.
Therefore,$P(A) = \frac{1}{4}$.

(D) The total number of possible outcomes when throwing a balanced die once is $n = 6$,which are $\{1, 2, 3, 4, 5, 6\}$.
Let $A$ be the event of getting an even number. The favourable outcomes for event $A$ are $\{2, 4, 6\}$.
Thus,the number of favourable outcomes is $m = 3$.
The probability of event $A$ is given by $P(A) = \frac{m}{n} = \frac{3}{6} = \frac{1}{2}$.

(C) standard balanced die has $6$ faces numbered from $1$ to $6$.
Since the number $7$ is not present on any face of the die,it is impossible to obtain $7$ in a single throw.
An event that cannot occur is called an impossible event,and its probability is always $0$.
Therefore,the required probability is $0$.

(C) The total number of cards in a well-shuffled pack is $n = 52$.
$A$ picture card (face card) is defined as a King,Queen,or Jack. There are $4$ Kings,$4$ Queens,and $4$ Jacks in a deck.
Therefore,the total number of favourable outcomes for drawing a picture card is $m = 4 + 4 + 4 = 12$.
The probability $P(A)$ of drawing a picture card is given by the ratio of favourable outcomes to the total number of outcomes:
$P(A) = \frac{m}{n} = \frac{12}{52} = \frac{3}{13}$.

(B) The total number of possible outcomes in a well-shuffled pack of $52$ cards is $n = 52$.
Let $A$ be the event that the card drawn is an ace.
Since there are $4$ aces in a standard deck of $52$ cards,the number of favourable outcomes is $m = 4$.
The probability of an event is given by the formula $P(A) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} = \frac{m}{n}$.
Therefore,$P(A) = \frac{4}{52} = \frac{1}{13}$.

(C) The total number of possible outcomes when drawing a card from a well-shuffled pack of $52$ cards is $n = 52$.
$A$ standard deck of cards contains two black suits: Spades and Clubs.
Each suit has $13$ cards,so the total number of black cards is $13 + 13 = 26$.
Let $A$ be the event that the card drawn is a black suit card.
The number of favourable outcomes for event $A$ is $m = 26$.
The probability $P(A)$ is given by the ratio of favourable outcomes to total outcomes:
$P(A) = \frac{m}{n} = \frac{26}{52} = \frac{1}{2}$.

(D) The total number of cards in a well-shuffled pack is $n = 52$.
$A$ picture card is defined as a King,Queen,or Jack. There are $4$ of each in a deck,so the total number of picture cards is $3 \times 4 = 12$.
Let $A$ be the event that the card drawn is not a picture card.
The number of favourable outcomes $m$ is the total number of cards minus the number of picture cards:
$m = 52 - 12 = 40$.
The probability $P(A)$ is given by the ratio of favourable outcomes to total outcomes:
$P(A) = \frac{m}{n} = \frac{40}{52}$.
Simplifying the fraction by dividing both numerator and denominator by $4$:
$P(A) = \frac{10}{13}$.

(C) In a standard pack of $52$ cards,there are $4$ suits,each containing $13$ cards.
The number of Diamond cards is $13$.
The number of cards that are not Diamonds is $52 - 13 = 39$.
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$\text{Probability (not a Diamond)} = \frac{\text{Number of non-Diamond cards}}{\text{Total number of cards}} = \frac{39}{52}$.
Simplifying the fraction by dividing both numerator and denominator by $13$,we get $\frac{39 \div 13}{52 \div 13} = \frac{3}{4}$.

(B) In a standard pack of $52$ cards,there are $4$ suits: Hearts,Diamonds,Clubs,and Spades.
Hearts and Diamonds are red suits,while Clubs and Spades are black suits.
Each suit contains exactly one ace.
Therefore,the number of aces of red suits is $2$ (one ace of Hearts and one ace of Diamonds).
The total number of possible outcomes is $52$.
The number of favorable outcomes is $2$.
Probability $= \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{2}{52} = \frac{1}{26}$.

(A) In a standard pack of $52$ cards,there are $4$ aces.
Therefore,the number of cards that are not aces is $52 - 4 = 48$.
The probability of drawing a card that is not an ace is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
$\text{Probability} = \frac{\text{Number of non-ace cards}}{\text{Total number of cards}} = \frac{48}{52}$.
Simplifying the fraction by dividing both numerator and denominator by $4$,we get $\frac{48 \div 4}{52 \div 4} = \frac{12}{13}$.

(C) The number of outcomes when throwing a balanced die once is $6$ (i.e.,$1, 2, 3, 4, 5, 6$).
When the die is thrown a second time,each of the $6$ results from the first throw can be paired with any of the $6$ results of the second throw.
Therefore,the total number of outcomes in the experiment of throwing a balanced die twice is $6 \times 6 = 36$.

(D) When one balanced die is rolled,the possible outcomes are ${1, 2, 3, 4, 5, 6}$,which means there are $6$ possible outcomes.
When two balanced dice are rolled simultaneously,the total number of outcomes is calculated by multiplying the number of outcomes of each die.
Total number of outcomes $= 6 \times 6 = 36$.

(A) When two balanced dice are rolled simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
Let $E$ be the event that the sum of the numbers on the two dice is $2$.
The only outcome that results in a sum of $2$ is $(1, 1)$.
Therefore,the number of favourable outcomes is $m = 1$.
The probability of the event $E$ is given by the formula: $P(E) = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$.
Thus,$P(E) = \frac{1}{36}$.

(C) When two balanced dice are rolled simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
Let $A$ be the event that the sum of the numbers on the two dice is $4$.
The favorable outcomes for this event are $(1, 3), (2, 2),$ and $(3, 1)$.
Thus,the number of favorable outcomes is $m = 3$.
The probability of the event $A$ is given by the formula $P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$.
$P(A) = \frac{3}{36} = \frac{1}{12}$.