Probability Questions and Answers

Q: Brother and sister both appear for an interview. The probability of the selection of brother is \(\frac{1}{6}\), while the probability of rejection of sister is \(\frac{3}{4}\). What is the probability that one of them is selected?

The probability is \(\frac{1}{3}\).The brother is selected with probability \(\frac{1}{6}\) (rejected \(\frac{5}{6}\)); the sister is rejected with probability \(\frac{3}{4}\), so selected with \(\frac{1}{4}\). Exactly one is selected if the brother is in and the sister out, or the sister in and the brother out:\[ \begin{aligned} \tfrac{1}{6} \times \tfrac{3}{4} + \tfrac{5}{6} \times \tfrac{1}{4} &= \tfrac{3}{24} + \tfrac{5}{24} \\ &= \tfrac{8}{24} = \tfrac{1}{3} \end{aligned} \]\(\frac{1}{4}\) is only the sister's chance, and \(\frac{5}{6}\) only the brother's chance of rejection.

Q: Cards bearing numbers 12, 13, 14, ... , 70, 71 (one number on one card) are kept in a box and then a card is drawn from the box without looking into it. What is the probability that the card drawn has the number which is not a prime ?

The probability is \(\frac{3}{4}\).The cards run from 12 to 71, so there are 71 - 12 + 1 = 60 cards. The primes among them are 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67 and 71, which is 15 numbers.\[ \begin{aligned} P(\text{not prime}) &= \frac{60 - 15}{60} \\ &= \frac{45}{60} = \frac{3}{4} \end{aligned} \]The options with denominator 59 miscount the cards (forgetting to include both ends), and \(\frac{1}{4}\) is the probability of drawing a prime.

Q: A coin is tossed 10 times and the outcomes are observed as :H, T, H, T, T, H, H, T, H, H(H is Head; T is Tail)What is the probability of getting Head?

The probability of getting a head is 3/5.This is an experimental probability, the number of times the event happened divided by the number of trials. Counting the heads in H, T, H, T, T, H, H, T, H, H gives 6 heads in 10 tosses:\[ P(\text{Head}) = \frac{6}{10} = \frac{3}{5} \]2/5 is the probability of a tail (4 tails out of 10), the opposite event. 4/5 and 1/5 do not match either count.

Q: Numbers from 8 to 92 were written on paper slips (one number on one slip) and were kept in a box. Then, a slip was taken out from the box, without looking into it. What is the probability that the number on the slip will be a multiple of 7 ?

The probability is \(\dfrac{12}{85}\).The slips carry the whole numbers from 8 to 92 inclusive:\[ 92 - 8 + 1 = 85 \text{ slips} \]The multiples of 7 in that range run from \(14 = 7 \times 2\) to \(91 = 7 \times 13\), which is \(13 - 2 + 1 = 12\) numbers. So\[ P = \frac{12}{85} \]\(\frac{13}{85}\) wrongly counts 7, which is below 8. \(\frac{13}{84}\) also miscounts the slips as \(92 - 8\), leaving out one end. \(\frac{1}{7}\) assumes exactly one number in seven, which is not true for this range.

Q: In which of the following situations do the results have an equal chance of occurring ?

The equally likely situation is captains of two teams tossing a coin to decide which team will bat first.A fair coin has two outcomes, head and tail, each with probability 1/2, so each captain has the same chance of winning the toss.The others are not equally likely. Whether a train is on time or late depends on many conditions, and the two outcomes need not have equal chances. A waitlisted ticket's chance of confirmation depends on its position and cancellations. A student who prepared 3 chapters out of 5 has a 3/5 chance of getting a question from them, not 1/2.

Q: A card is drawn from a well-shuffled deck of playing cards. What is the probability that the card drawn is a 10 or a red king?

The probability is \(\frac{3}{26}\).A deck of 52 cards has four 10s and two red kings (hearts and diamonds). These cards do not overlap, so the favourable outcomes simply add.\[ P = \frac{4 + 2}{52} = \frac{6}{52} = \frac{3}{26} \]\(\frac{1}{13}\) counts only the 10s, \(\frac{1}{26}\) counts only the red kings, and \(\frac{2}{13}\) counts all four kings instead of the two red ones.

Q: A bag contains 3 white, 2 blue and 5 red balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is not red ?

The probability that the ball is not red is \(\frac{1}{2}\).There are 3 + 2 + 5 = 10 balls in all, of which 5 are red, so 5 are not red (3 white and 2 blue):\[ P(\text{not red}) = \frac{5}{10} = \frac{1}{2} \]Equivalently, \(1 - P(\text{red}) = 1 - \frac{5}{10}\). \(\frac{3}{10}\) is the probability of white alone, \(\frac{1}{5}\) is the probability of blue alone, and \(\frac{4}{5}\) does not correspond to any event here.

Q: Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict is

The probability that B hits is about 0.14.Let B hit with probability p, so A hits with probability 3p. The shots are independent, and the convict escapes injury only if both miss:\[ \begin{aligned} (1 - 3p)(1 - p) &= 0.5 \\ 3p^{2} - 4p + 0.5 &= 0 \\ p &= \frac{4 - \sqrt{10}}{6} \approx 0.14 \end{aligned} \]The other root, about 1.19, is not a probability. Check: A hits with 0.42, and \(0.58 \times 0.86 \approx 0.50\). The value 0.33 would make A's probability 0.99, and 0.22 or 0.40 give a non-injury probability far from 0.5.

Q: Shaquille O' Neal is a 60% career free throw shooter, meaning that he successfully makes 60 free throws out of 100 attempts on average. What is the probability that he will successfully make exactly 6 free throws in 10 attempts?

The probability is about 0.2508.Each attempt succeeds with probability 0.6, independently, so the number of successes in 10 attempts is binomial:\[ \begin{aligned} P(6) &= \binom{10}{6}(0.6)^{6}(0.4)^{4} \\ &= 210 \times 0.046656 \times 0.0256 \\ &\approx 0.2508 \end{aligned} \]0.6000 is the success rate per attempt, not the chance of exactly 6 in 10. The other values come from errors in the binomial coefficient or the powers; even the most likely count of a binomial has probability well below 0.6.

Q: An unbiased coin is tossed six times in a row and four different such trials are conducted. One trial implies six tosses of the coin. If H stands for head and T stands for tail, the following are the observations from the four trials:(1) HTHTHT (2) TTHHHT (3) HTTHHT (4) HHHT__ __.Which statement describing the last two coin tosses of the fourth trial has the highest probability of being correct?

The most likely statement is One H and one T will occur.Tosses of an unbiased coin are independent, so the earlier trials and the first four tosses of the fourth trial tell us nothing about the last two. Those two tosses have four equally likely outcomes: HH, HT, TH and TT, each with probability 1/4.'One H and one T' is satisfied by both HT and TH, so its probability is 1/2. 'Two T' (TT only) and 'Two H' (HH only) each have 1/4. 'One H followed by one T' fixes the order, so it covers HT alone and also has 1/4.

Important Instructions

Brother and sister both appear for an interview. The probability of the selection of brother is \(\frac{1}{6}\), while the probability of rejection of sister is \(\frac{3}{4}\). What is the probability that one of them is selected?

A.\(\frac{2}{3}\)
B.\(\frac{1}{3}\)
C.\(\frac{1}{4}\)
D.\(\frac{5}{6}\)

Exam Year(s)

CBSE Assistant Secretary - 2024,

Answer: B

The probability is \(\frac{1}{3}\).

The brother is selected with probability \(\frac{1}{6}\) (rejected \(\frac{5}{6}\)); the sister is rejected with probability \(\frac{3}{4}\), so selected with \(\frac{1}{4}\). Exactly one is selected if the brother is in and the sister out, or the sister in and the brother out:

\[ \begin{aligned} \tfrac{1}{6} \times \tfrac{3}{4} + \tfrac{5}{6} \times \tfrac{1}{4} &= \tfrac{3}{24} + \tfrac{5}{24} \\ &= \tfrac{8}{24} = \tfrac{1}{3} \end{aligned} \]

\(\frac{1}{4}\) is only the sister's chance, and \(\frac{5}{6}\) only the brother's chance of rejection.

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Cards bearing numbers 12, 13, 14, ... , 70, 71 (one number on one card) are kept in a box and then a card is drawn from the box without looking into it. What is the probability that the card drawn has the number which is not a prime ?

A.\(\frac{45}{59}\)
B.\(\frac{1}{4}\)
C.\(\frac{15}{59}\)
D.\(\frac{3}{4}\)

Exam Year(s)

CTET - 2024,

Answer: D

The probability is \(\frac{3}{4}\).

The cards run from 12 to 71, so there are 71 - 12 + 1 = 60 cards. The primes among them are 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67 and 71, which is 15 numbers.

\[ \begin{aligned} P(\text{not prime}) &= \frac{60 - 15}{60} \\ &= \frac{45}{60} = \frac{3}{4} \end{aligned} \]

The options with denominator 59 miscount the cards (forgetting to include both ends), and \(\frac{1}{4}\) is the probability of drawing a prime.

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A coin is tossed 10 times and the outcomes are observed as :

H, T, H, T, T, H, H, T, H, H
(H is Head; T is Tail)

What is the probability of getting Head?

A.\(\frac{3}{5}\)
B.\(\frac{4}{5}\)
C.\(\frac{2}{5}\)
D.\(\frac{1}{5}\)

Exam Year(s)

CTET - 2018,

Answer: A

The probability of getting a head is 3/5.

This is an experimental probability, the number of times the event happened divided by the number of trials. Counting the heads in H, T, H, T, T, H, H, T, H, H gives 6 heads in 10 tosses:

\[ P(\text{Head}) = \frac{6}{10} = \frac{3}{5} \]

2/5 is the probability of a tail (4 tails out of 10), the opposite event. 4/5 and 1/5 do not match either count.

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Numbers from 8 to 92 were written on paper slips (one number on one slip) and were kept in a box. Then, a slip was taken out from the box, without looking into it. What is the probability that the number on the slip will be a multiple of 7 ?

A.\(\dfrac{12}{85}\)
B.\(\dfrac{13}{84}\)
C.\(\dfrac{1}{7}\)
D.\(\dfrac{13}{85}\)

Exam Year(s)

CTET - 2026,

Answer: A

The probability is \(\dfrac{12}{85}\).

The slips carry the whole numbers from 8 to 92 inclusive:

\[ 92 - 8 + 1 = 85 \text{ slips} \]

The multiples of 7 in that range run from \(14 = 7 \times 2\) to \(91 = 7 \times 13\), which is \(13 - 2 + 1 = 12\) numbers. So

\[ P = \frac{12}{85} \]

\(\frac{13}{85}\) wrongly counts 7, which is below 8. \(\frac{13}{84}\) also miscounts the slips as \(92 - 8\), leaving out one end. \(\frac{1}{7}\) assumes exactly one number in seven, which is not true for this range.

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In which of the following situations do the results have an equal chance of occurring ?

A.The chances of a train being on time or late.
B.The chances of a ticket which was waitlisted being confirmed.
C.A student getting a question from the 3 chapters she prepared, out of 5.
D.Captains of two teams tossing a coin to decide which team will bat first.

Exam Year(s)

CTET - 2026,

Answer: D

The equally likely situation is captains of two teams tossing a coin to decide which team will bat first.

A fair coin has two outcomes, head and tail, each with probability 1/2, so each captain has the same chance of winning the toss.

The others are not equally likely. Whether a train is on time or late depends on many conditions, and the two outcomes need not have equal chances. A waitlisted ticket's chance of confirmation depends on its position and cancellations. A student who prepared 3 chapters out of 5 has a 3/5 chance of getting a question from them, not 1/2.

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A card is drawn from a well-shuffled deck of playing cards. What is the probability that the card drawn is a 10 or a red king?

A.\(\frac{1}{26}\)
B.\(\frac{1}{13}\)
C.\(\frac{3}{26}\)
D.\(\frac{2}{13}\)

Exam Year(s)

CTET - 2026,

Answer: C

The probability is \(\frac{3}{26}\).

A deck of 52 cards has four 10s and two red kings (hearts and diamonds). These cards do not overlap, so the favourable outcomes simply add.

\[ P = \frac{4 + 2}{52} = \frac{6}{52} = \frac{3}{26} \]

\(\frac{1}{13}\) counts only the 10s, \(\frac{1}{26}\) counts only the red kings, and \(\frac{2}{13}\) counts all four kings instead of the two red ones.

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A bag contains 3 white, 2 blue and 5 red balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is not red ?

A.\(\frac{4}{5}\)
B.\(\frac{3}{10}\)
C.\(\frac{1}{5}\)
D.\(\frac{1}{2}\)

Exam Year(s)

CTET - 2019,

Answer: D

The probability that the ball is not red is \(\frac{1}{2}\).

There are 3 + 2 + 5 = 10 balls in all, of which 5 are red, so 5 are not red (3 white and 2 blue):

\[ P(\text{not red}) = \frac{5}{10} = \frac{1}{2} \]

Equivalently, \(1 - P(\text{red}) = 1 - \frac{5}{10}\). \(\frac{3}{10}\) is the probability of white alone, \(\frac{1}{5}\) is the probability of blue alone, and \(\frac{4}{5}\) does not correspond to any event here.

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Two policemen, A and B, fire once each at the same time at an escaping convict. The probability that A hits the convict is three times the probability that B hits the convict. If the probability of the convict not getting injured is 0.5, the probability that B hits the convict is

A.0.14
B.0.22
C.0.33
D.0.40

Exam Year(s)

GATE AE - 2012, GATE AG - 2012,

Answer: A

The probability that B hits is about 0.14.

Let B hit with probability p, so A hits with probability 3p. The shots are independent, and the convict escapes injury only if both miss:

\[ \begin{aligned} (1 - 3p)(1 - p) &= 0.5 \\ 3p^{2} - 4p + 0.5 &= 0 \\ p &= \frac{4 - \sqrt{10}}{6} \approx 0.14 \end{aligned} \]

The other root, about 1.19, is not a probability. Check: A hits with 0.42, and \(0.58 \times 0.86 \approx 0.50\). The value 0.33 would make A's probability 0.99, and 0.22 or 0.40 give a non-injury probability far from 0.5.

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Shaquille O' Neal is a 60% career free throw shooter, meaning that he successfully makes 60 free throws out of 100 attempts on average. What is the probability that he will successfully make exactly 6 free throws in 10 attempts?

A.0.2508
B.0.2816
C.0.2934
D.0.6000

Exam Year(s)

GATE AE - 2016,

Answer: A

The probability is about 0.2508.

Each attempt succeeds with probability 0.6, independently, so the number of successes in 10 attempts is binomial:

\[ \begin{aligned} P(6) &= \binom{10}{6}(0.6)^{6}(0.4)^{4} \\ &= 210 \times 0.046656 \times 0.0256 \\ &\approx 0.2508 \end{aligned} \]

0.6000 is the success rate per attempt, not the chance of exactly 6 in 10. The other values come from errors in the binomial coefficient or the powers; even the most likely count of a binomial has probability well below 0.6.

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10.

An unbiased coin is tossed six times in a row and four different such trials are conducted. One trial implies six tosses of the coin. If H stands for head and T stands for tail, the following are the observations from the four trials:

(1) HTHTHT (2) TTHHHT (3) HTTHHT (4) HHHT__ __.

Which statement describing the last two coin tosses of the fourth trial has the highest probability of being correct?

A.Two T will occur.
B.One H and one T will occur.
C.Two H will occur.
D.One H will be followed by one T.

Exam Year(s)

GATE Mathematics - 2018, GATE ME - 2018, GATE AE - 2018,

Answer: B

The most likely statement is One H and one T will occur.

Tosses of an unbiased coin are independent, so the earlier trials and the first four tosses of the fourth trial tell us nothing about the last two. Those two tosses have four equally likely outcomes: HH, HT, TH and TT, each with probability 1/4.

'One H and one T' is satisfied by both HT and TH, so its probability is 1/2. 'Two T' (TT only) and 'Two H' (HH only) each have 1/4. 'One H followed by one T' fixes the order, so it covers HT alone and also has 1/4.

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