axiomatic-probability MCQs for UPSC Prelims

35 practice questions on axiomatic-probability from the Probability section of the UPSC Prelims syllabus. 35 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

11 Easy 17 Medium 7 Hard

Sample questions

Q1
medium

A bag holds 6 white balls, 5 black balls and 4 yellow balls, alike apart from colour. One ball is drawn at random. What is the probability that the ball drawn is white or yellow?

  1. A $\frac{11}{15}$
  2. B $\frac{1}{3}$
  3. C $\frac{2}{5}$
  4. D $\frac{2}{3}$
Show answer and explanation

Correct answer: D - $\frac{2}{3}$

The bag holds $6 + 5 + 4 = 15$ equally likely balls, and one ball cannot be both white and yellow, so the two probabilities add: $\frac{6}{15} + \frac{4}{15} = \frac{10}{15} = \frac{2}{3}$. The value $\frac{1}{3}$ is the chance of a black ball. The value $\frac{2}{5}$ counts the white balls alone, and $\frac{11}{15}$ adds the black balls in place of the yellow ones.

Q2
easy

Following the axioms of probability, which statement holds for every event $E$ of a sample space $S$?

  1. A $P(E) > 0$
  2. B $0 \leq P(E) \leq 1$
  3. C $P(E) \geq 1$
  4. D $P(E)$ may be any real number
Show answer and explanation

Correct answer: B - $0 \leq P(E) \leq 1$

The axioms demand $P(E) \geq 0$ for every event together with $P(S) = 1$, and since $E$ is part of $S$ its probability cannot rise above 1, so every probability sits between 0 and 1. Strict positivity fails for the impossible event, whose probability is 0. Values below 0 or above 1 are forbidden outright, so the last two choices are wrong.

Q3
hard

Let $A$ and $B$ be events of one sample space with $P(A) = 0.7$, $P(B) = 0.6$ and $P(A \cap B) = 0.4$. Assertion (A): $P(A \cup B) = 0.9$. Reason (R): For any two events, $P(A \cup B) = P(A) + P(B)$.

  1. A Both A and R are true, and R is the correct explanation of A
  2. B Both A and R are true, but R is not the correct explanation of A
  3. C A is true, but R is false
  4. D A is false, but R is true
Show answer and explanation

Correct answer: C - A is true, but R is false

The addition rule gives $P(A \cup B) = 0.7 + 0.6 - 0.4 = 0.9$, so A is true. R is false as a general statement, because the plain sum holds only for mutually exclusive events, and here the overlap has probability $0.4$. Following R alone would give $1.3$, which is impossible for a probability.

Q4
easy

A box holds 20 cards of the same size numbered 1 to 20, and one card is drawn at random. What is the probability that the number on it is a multiple of 5?

  1. A $\frac{1}{5}$
  2. B $\frac{1}{4}$
  3. C $\frac{3}{20}$
  4. D $\frac{1}{20}$
Show answer and explanation

Correct answer: A - $\frac{1}{5}$

The multiples of 5 from 1 to 20 are 5, 10, 15 and 20, so 4 of the 20 equally likely cards are favourable and the probability is $\frac{4}{20} = \frac{1}{5}$. The value $\frac{1}{4}$ counts five favourable cards, as though 25 were in the box. The value $\frac{1}{20}$ counts a single card, and $\frac{3}{20}$ leaves the card 20 out of the list.

Q5
medium

One card is drawn at random from a well-shuffled standard pack of 52 playing cards. Assertion (A): The events 'the card is a king' and 'the card is a heart' are mutually exclusive. Reason (R): A standard pack of 52 playing cards has four suits of 13 cards each.

  1. A Both A and R are true, and R is the correct explanation of A
  2. B Both A and R are true, but R is not the correct explanation of A
  3. C A is true, but R is false
  4. D A is false, but R is true
Show answer and explanation

Correct answer: D - A is false, but R is true

R is true, since the pack is built from four suits of 13 cards. A is false, because the king of hearts is a king and a heart at the same time, so the two events share an outcome and cannot be mutually exclusive. Their overlap is that one card, which is why $P(\text{king or heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{4}{13}$.

Q6
medium

Three coins that can be told apart are tossed together once. What is the probability of getting at most two heads?

  1. A $\frac{1}{2}$
  2. B $\frac{1}{8}$
  3. C $\frac{3}{8}$
  4. D $\frac{7}{8}$
Show answer and explanation

Correct answer: D - $\frac{7}{8}$

Among the eight equally likely records, 'at most two heads' leaves out only $HHH$, so it is the complement of 'three heads' and its probability is $1 - \frac{1}{8} = \frac{7}{8}$. The value $\frac{1}{8}$ is the chance of three heads. The value $\frac{3}{8}$ counts exactly two heads only, and $\frac{1}{2}$ is the chance of a head on one chosen coin.

Q7
medium

The odds in favour of an event $E$ are 3 to 7, which means that out of 10 equally likely chances, 3 favour $E$ and the other 7 do not. What is $P(E')$?

  1. A $\frac{3}{7}$
  2. B $\frac{7}{3}$
  3. C $\frac{3}{10}$
  4. D $\frac{7}{10}$
Show answer and explanation

Correct answer: D - $\frac{7}{10}$

Of the 10 equally likely chances, 3 favour $E$, so $P(E) = \frac{3}{10}$ and the complement takes the remaining 7 chances, giving $P(E') = \frac{7}{10}$. The value $\frac{3}{10}$ is $P(E)$ itself, not its complement. The value $\frac{3}{7}$ turns the odds ratio into a fraction without using the total of 10 chances, and $\frac{7}{3}$ is more than 1, so it cannot be a probability.

Q8
easy

For an event $A$ of a sample space, which relation connects $P(A)$ with the probability of its complement $A'$?

  1. A $P(A') = \frac{1}{P(A)}$
  2. B $P(A') = P(A) - 1$
  3. C $P(A') = 1 + P(A)$
  4. D $P(A') = 1 - P(A)$
Show answer and explanation

Correct answer: D - $P(A') = 1 - P(A)$

An event and its complement are mutually exclusive and together make up the whole sample space, so $P(A) + P(A') = P(S) = 1$, which gives $P(A') = 1 - P(A)$. The second choice is negative whenever $P(A) < 1$. The third can exceed 1, and the reciprocal is not a probability at all - for $P(A) = 0.2$ it would give 5.

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