bayes-theorem MCQs for UPSC Prelims

36 practice questions on bayes-theorem from the Probability section of the UPSC Prelims syllabus. 36 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

10 Easy 18 Medium 8 Hard

Sample questions

Q1
easy

Let $E_1,E_2,\dots,E_n$ be a partition of the sample space with every $P(E_i)>0$, and let $A$ be an event with $P(A)>0$. Bayes' theorem gives $P(E_i\mid A)$ as

  1. A $P(E_i)\,P(A\mid E_i)$
  2. B $\frac{P(A\mid E_i)}{\sum_{j} P(A\mid E_j)}$
  3. C $\frac{P(E_i)\,P(A\mid E_i)}{\sum_{j} P(E_j)\,P(A\mid E_j)}$
  4. D $\frac{P(E_i)\,P(A\mid E_i)}{P(E_i)}$
Show answer and explanation

Correct answer: C - $\frac{P(E_i)\,P(A\mid E_i)}{\sum_{j} P(E_j)\,P(A\mid E_j)}$

By definition $P(E_i\mid A)=\frac{P(E_i\cap A)}{P(A)}$; the multiplication theorem turns the numerator into $P(E_i)\,P(A\mid E_i)$ and the theorem of total probability expands the denominator over all branches. Dropping the priors leaves a ratio of likelihoods that ignores how likely each branch was to begin with. Dividing by $P(E_i)$ merely returns $P(A\mid E_i)$. The bare product is $P(E_i\cap A)$, never rescaled by $P(A)$.

Q2
medium

Let $E$ be an event with $0<P(E)<1$ and let $F$ be any other event. Which of the following always forms a partition of the sample space?

  1. A $E\cap F$ and $E\cup F$
  2. B $E$ and $E\cup F$
  3. C $E$ and $S$
  4. D $E$ and $E'$
Show answer and explanation

Correct answer: D - $E$ and $E'$

The pair $E$ and $E'$ is disjoint, covers the whole sample space, and since $0<P(E)<1$ both members carry non-zero probability, so it is a partition. Pairing $E$ with $S$ fails because $E$ lies inside $S$. The intersection $E\cap F$ lies inside the union $E\cup F$, so that pair overlaps too. Finally $E$ lies inside $E\cup F$, which again breaks disjointness.

Q3
medium

A man speaks the truth on $4$ out of every $5$ occasions. He throws a fair die once and, on being asked whether it showed a six, he answers 'yes'. Find the probability that the die really did show a six.

  1. A $\frac{4}{9}$
  2. B $\frac{5}{9}$
  3. C $\frac{1}{6}$
  4. D $\frac{4}{5}$
Show answer and explanation

Correct answer: A - $\frac{4}{9}$

He answers 'yes' either when the die shows a six and he tells the truth, with probability $\frac{1}{6}\times\frac{4}{5}=\frac{4}{30}$, or when it does not and he lies, with probability $\frac{5}{6}\times\frac{1}{5}=\frac{5}{30}$. The total is $\frac{9}{30}$, so the answer is $\frac{4/30}{9/30}=\frac{4}{9}$. The value $\frac{5}{9}$ is the posterior that the die did not show a six. The value $\frac{4}{5}$ is his truthfulness, the reversed conditional, and $\frac{1}{6}$ is the prior.

Q4
hard

Three identical boxes are given. Box I holds $1$ gold and $4$ silver coins, Box II holds $3$ gold and $2$ silver coins, and Box III holds $4$ gold and $1$ silver coin. A box is chosen at random and a coin drawn from it is gold. Find the probability that it came from Box II.

  1. A $\frac{3}{5}$
  2. B $\frac{1}{2}$
  3. C $\frac{3}{8}$
  4. D $\frac{1}{8}$
Show answer and explanation

Correct answer: C - $\frac{3}{8}$

Each box has prior $\frac{1}{3}$, so the total probability of a gold coin is $\frac{1}{3}\left(\frac{1}{5}+\frac{3}{5}+\frac{4}{5}\right)=\frac{8}{15}$, while Box II contributes $\frac{1}{3}\times\frac{3}{5}=\frac{3}{15}$. The answer is $\frac{3/15}{8/15}=\frac{3}{8}$. The values $\frac{1}{8}$ and $\frac{1}{2}$ are the posteriors for Box I and Box III, and the three add to $1$. The value $\frac{3}{5}$ is the gold proportion in Box II, the reversed conditional.

Q5
medium

The events $E_1$, $E_2$ and $E_3$ form a partition of a sample space with $P(E_1)=\frac{1}{6}$, $P(E_2)=\frac{1}{3}$ and $P(E_3)=\frac{1}{2}$. For an event $A$, $P(A\mid E_1)=\frac{1}{2}$, $P(A\mid E_2)=\frac{1}{4}$ and $P(A\mid E_3)=\frac{1}{5}$. Find $P(A)$.

  1. A $\frac{1}{2}$
  2. B $\frac{4}{15}$
  3. C $\frac{19}{20}$
  4. D $\frac{19}{60}$
Show answer and explanation

Correct answer: B - $\frac{4}{15}$

Total probability weights each conditional by its own branch: $\frac{1}{6}\times\frac{1}{2}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{2}\times\frac{1}{5}=\frac{1}{12}+\frac{1}{12}+\frac{1}{10}=\frac{4}{15}$. Adding the conditionals alone gives $\frac{19}{20}$ and averaging them gives $\frac{19}{60}$; both discard the unequal weights. The value $\frac{1}{2}$ is simply the largest of the three conditionals.

Q6
medium

Assertion (A): Two bags are equally likely to be chosen. Bag I holds $1$ white and $4$ black balls, and Bag II holds $3$ white and $2$ black balls. If the ball drawn from the chosen bag is white, the probability that Bag II was the one chosen is $\frac{3}{4}$. Reason (R): When the prior probabilities are equal, each posterior probability is proportional to that branch's chance of producing the observed outcome.

  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: A - Both A and R are true, and R is the correct explanation of A

The total probability of a white ball is $\frac{1}{2}\times\frac{1}{5}+\frac{1}{2}\times\frac{3}{5}=\frac{2}{5}$, and Bag II supplies $\frac{3}{10}$ of it, giving $\frac{3/10}{2/5}=\frac{3}{4}$, so (A) is true. (R) is true as well: with the common factor $\frac{1}{2}$ cancelling from every term, the posteriors fall in the ratio $1:3$ of the two white-ball proportions, which is exactly how the value in (A) arises.

Q7
medium

Three identical urns are placed on a table. Urn I holds $2$ white and $3$ black balls, Urn II holds $4$ white and $1$ black ball, and Urn III holds $3$ white and $2$ black balls. An urn is chosen at random and one ball is drawn from it. Find the probability that the ball is white.

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

Correct answer: C - $\frac{3}{5}$

Each urn is chosen with probability $\frac{1}{3}$, so $P(\text{white})=\frac{1}{3}\left(\frac{2}{5}+\frac{4}{5}+\frac{3}{5}\right)=\frac{1}{3}\times\frac{9}{5}=\frac{3}{5}$. The value $\frac{1}{3}$ is the chance of picking a particular urn and says nothing about colour. The values $\frac{4}{5}$ and $\frac{2}{5}$ are the white proportions of Urn II and Urn I, each used as though that urn were certain to be chosen.

Q8
medium

Assertion (A): In Bayes' theorem the denominator is the total probability of the observed event. Reason (R): Bayes' theorem can be applied only when the partition consists of exactly two events.

  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

(A) is true: the denominator adds $P(E_j)\,P(A\mid E_j)$ over every branch, which is precisely $P(A)$. (R) is false, because the theorem holds for a partition into any finite number of events of non-zero probability; the two-branch case built from $E$ and $E'$ is only the commonest example. Extra branches simply add more terms to that denominator.

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