standard-experiments MCQs for UPSC Prelims

33 practice questions on standard-experiments from the Probability section of the UPSC Prelims syllabus. 33 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

10 Easy 16 Medium 7 Hard

Sample questions

Q1
medium

Assertion (A): When two fair coins are tossed together, the probability of getting at least one head is $\frac{3}{4}$. Reason (R): The four outcomes HH, HT, TH and TT of tossing two coins are equally likely.

  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

Three of the four outcomes, namely HH, HT and TH, contain at least one head, so the probability is $\frac{3}{4}$ and the assertion is true. The reason is a true statement about the sample space. It is exactly because those four outcomes are equally likely that counting three of them gives $\frac{3}{4}$, so R explains A.

Q2
medium

A wheel has $10$ equal sectors numbered $1$ to $10$, and the pointer is equally likely to stop on any sector. What is the probability that it does not stop on a number greater than $7$?

  1. A $\frac{3}{10}$
  2. B $\frac{7}{10}$
  3. C $\frac{1}{2}$
  4. D $\frac{4}{5}$
Show answer and explanation

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

The numbers greater than $7$ are $8$, $9$ and $10$, so that event has probability $\frac{3}{10}$ and its complement has probability $1-\frac{3}{10}=\frac{7}{10}$. The value $\frac{3}{10}$ is the event itself rather than its complement. The value $\frac{4}{5}$ treats only $9$ and $10$ as greater than $7$. The value $\frac{1}{2}$ splits the wheel at $5$ instead of at $7$.

Q3
hard

A wheel is divided into $20$ equal sectors numbered $1$ to $20$, and the pointer is equally likely to stop on any sector. What is the probability that it stops on a prime number?

  1. A $\frac{1}{2}$
  2. B $\frac{7}{20}$
  3. C $\frac{9}{20}$
  4. D $\frac{2}{5}$
Show answer and explanation

Correct answer: D - $\frac{2}{5}$

The primes up to $20$ are $2$, $3$, $5$, $7$, $11$, $13$, $17$ and $19$, that is $8$ numbers, so the probability is $\frac{8}{20}=\frac{2}{5}$. The value $\frac{9}{20}$ counts $1$ as a prime, which it is not. The value $\frac{1}{2}$ counts all ten odd numbers as prime, but $1$, $9$ and $15$ are not. The value $\frac{7}{20}$ drops $2$, the only even prime.

Q4
hard

Two dice, one yellow and one blue, are thrown together. What is the probability that the product of the two numbers is a multiple of $5$?

  1. A $\frac{2}{7}$
  2. B $\frac{11}{36}$
  3. C $\frac{1}{3}$
  4. D $\frac{1}{6}$
Show answer and explanation

Correct answer: B - $\frac{11}{36}$

A product of two numbers from $1$ to $6$ is a multiple of $5$ only when at least one die shows $5$. Six outcomes have $5$ on the yellow die and six have $5$ on the blue, and $(5,5)$ is in both lists, so $6+6-1=11$ outcomes are favourable, giving $\frac{11}{36}$. The value $\frac{1}{3}$ counts $(5,5)$ twice. The value $\frac{1}{6}$ uses only the yellow die. The value $\frac{2}{7}$ treats the dice as identical.

Q5
easy

A fair one rupee coin is tossed once. What is the probability that it shows a tail?

  1. A $\frac{1}{2}$
  2. B $0$
  3. C $\frac{1}{4}$
  4. D $1$
Show answer and explanation

Correct answer: A - $\frac{1}{2}$

A single toss has two equally likely outcomes, head and tail, and one of them is favourable, so the probability is $\frac{1}{2}$. The value $1$ would make a tail certain, which it is not. The value $\frac{1}{4}$ belongs to a pair of tosses rather than one toss. The value $0$ would make a tail impossible.

Q6
easy

A wheel is divided into $6$ equal sectors numbered $1$ to $6$, and the pointer is equally likely to stop on any sector. What is the probability that it stops on a number greater than $4$?

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

Correct answer: B - $\frac{1}{3}$

The sectors numbered more than $4$ are $5$ and $6$, so $2$ of the $6$ sectors are favourable and the probability is $\frac{2}{6}=\frac{1}{3}$. The value $\frac{2}{3}$ is the probability of stopping on $4$ or less. The value $\frac{1}{2}$ counts three sectors, as though $4$ were included. The value $\frac{1}{6}$ counts only one sector.

Q7
hard

All $12$ face cards are removed from a well-shuffled standard pack of $52$ playing cards. One card is then drawn at random from the cards that remain. What is the probability that it is red?

  1. A $\frac{5}{13}$
  2. B $\frac{1}{2}$
  3. C $\frac{13}{20}$
  4. D $\frac{1}{4}$
Show answer and explanation

Correct answer: B - $\frac{1}{2}$

Removing the $12$ face cards leaves $40$ cards, and $6$ of the removed cards were red, so $26-6=20$ red cards remain and the probability is $\frac{20}{40}=\frac{1}{2}$. The value $\frac{13}{20}$ keeps all $26$ red cards in the numerator. The value $\frac{5}{13}$ divides the $20$ red cards by the original $52$. The value $\frac{1}{4}$ counts only the $10$ hearts left, forgetting the diamonds.

Q8
medium

Two fair coins are tossed together. What is the probability of getting exactly one head?

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

Correct answer: C - $\frac{1}{2}$

Among the outcomes HH, HT, TH and TT, exactly one head appears in HT and TH, so the probability is $\frac{2}{4}=\frac{1}{2}$. The value $\frac{1}{4}$ counts only HT and forgets TH. The value $\frac{1}{3}$ treats the records $0$, $1$ and $2$ heads as equally likely. The value $\frac{3}{4}$ answers 'at least one head'.

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