Probability Questions and Answers

Q: In a standard deck of 52 cards, what is the probability of drawing a red card ?

The probability of a red card is 1/2.Half the deck is red: 13 hearts and 13 diamonds, 26 of the 52 cards. So P(red) = 26/52 = 1/2.1/4 is the probability of one particular suit, 1/13 of one particular rank, and 1/52 of one particular card.The same reasoning gives 1/2 for a black card, since the other two suits, spades and clubs, are black.

Q: If the odds in favour of an event are 3:4, what is the probability of the event occurring ?

The probability is 3/7.Odds in favour of 3 : 4 mean 3 favourable outcomes for every 4 unfavourable ones, 7 in all. So the probability of the event is 3/(3 + 4) = 3/7.4/7 is the probability of the event not happening, 3/4 is the odds written as a fraction, and 7/3 is greater than 1, which no probability can be.In general, odds of a : b in favour give a probability of a/(a + b).

Q: Which of the following is a characteristic of a binomial distribution ?

A binomial distribution has two outcomes, success or failure, on each trial.The binomial distribution counts successes in a fixed number n of independent trials, each with only two outcomes and the same probability of success p.The trials are independent, not dependent, and the count of successes is discrete, not continuous. The mean (np) need not equal the mode, so that is not a defining feature.

Q: Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty ?

The probability is \(pq + (1-p)(1-q)\).A computer is declared faulty in two ways:it is faulty (probability p) and the test is correct (probability q): \(pq\);it is good (probability 1 - p) and the test is wrong (probability 1 - q): \((1-p)(1-q)\).Adding the two disjoint cases gives the answer. \(pq\) alone counts only the correctly detected faults, while \((1-q)p\) and \((1-p)q\) are the cases where a faulty machine passes or a good machine is correctly passed.

Q: X is a uniformly distributed random variable that takes value between 0 and 1.The value of E(X3) will be

\(E(X^{3}) = \tfrac{1}{4}\).For X uniform on (0, 1) the density is 1, so\[ E(X^{3}) = \int_{0}^{1} x^{3}\,dx = \frac{1}{4} \]1/2 is \(E(X)\) and 1/8 is \(E(X)^{3}\), the cube of the mean, which is not the mean of the cube. Zero would need X to be symmetric about 0.In general, for X uniform on (0, 1), the mean of the k-th power is 1/(k + 1), so the second moment is 1/3 and the third 1/4.

Q: What is the probability of observing exactly k successes in n trials in a binomial distribution with parameter p ?

The probability is \({}^{n}C_{k}\, p^{k}(1-p)^{n-k}\).Any particular sequence of k successes and n - k failures has probability \(p^{k}(1-p)^{n-k}\), and there are \({}^{n}C_{k}\) ways of choosing which k of the n trials are the successes.Without the \({}^{n}C_{k}\) factor you get the probability of one particular sequence only. \(p(1-p)^{k}\) and \({}^{n}C_{k}\, p^{n}(1-p)^{k}\) put the exponents in the wrong places.

Q: Conditional Probability: In a class, 60% of the students are girls, and 70% of the students like math. If 80% of the girls like math, what is the probability that a randomly selected math-loving student is a girl ?

The probability is about 0.68.By Bayes' theorem, with G = girl and M = likes math:\[ \begin{aligned} P(G \mid M) &= \frac{P(M \mid G)\,P(G)}{P(M)} \\ &= \frac{0.8 \times 0.6}{0.7} = \frac{0.48}{0.7} \approx 0.686 \end{aligned} \]0.68 is the closest option. 0.48 would be the joint probability of a girl who likes math, which is not offered, and 0.25, 0.50 and 0.42 do not follow from the data.

Q: In a group of 100 students, 60 play cricket, 50 play football, and 30 play both. If a student is selected at random, what is the probability that the student plays cricket given that the student plays football ?

The probability is 0.60.Given that the student plays football, we look only at the 50 football players, of whom 30 also play cricket:\[ P(C \mid F) = \frac{30}{50} = 0.60 \]0.30 is the unconditional probability of playing both (30 out of 100), and 0.15 and 0.18 do not follow from the figures.The general rule is P(C | F) = P(C and F) / P(F) = 0.30 / 0.50, which gives the same 0.60.

Q: Find the variance in uniform distribution if X is uniformly distributed over (5, 30)?

The variance is 52.083.For a uniform distribution on (a, b) the variance is \((b-a)^{2}/12\):\[ \frac{(30 - 5)^{2}}{12} = \frac{625}{12} \approx 52.083 \]17.5 is the mean, \((5 + 30)/2\), not the variance, and 1/5 and 85.052 do not come from the formula.For any uniform distribution the mean is the midpoint (a + b)/2 and the variance depends only on the width b - a, so shifting the interval does not change the variance.

Q: Compute the Median for the exponential distribution with parameter λ :

The median is \(\log(2)/\lambda\).For an exponential distribution with rate \(\lambda\), \(P(X \le x) = 1 - e^{-\lambda x}\). The median m makes this one half:\[ \begin{aligned} 1 - e^{-\lambda m} &= \tfrac{1}{2} \\ m &= \frac{\ln 2}{\lambda} \end{aligned} \]Here log means the natural logarithm. \(\log(2)/2\lambda\) and \(2\log(2)/\lambda\) are off by a factor of two, and \(\log(3)/\lambda\) would be the point with probability 2/3 below it.

Important Instructions

41.

In a standard deck of 52 cards, what is the probability of drawing a red card ?

A.1/2
B.1/4
C.1/13
D.1/52

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

The probability of a red card is 1/2.

Half the deck is red: 13 hearts and 13 diamonds, 26 of the 52 cards. So P(red) = 26/52 = 1/2.

1/4 is the probability of one particular suit, 1/13 of one particular rank, and 1/52 of one particular card.

The same reasoning gives 1/2 for a black card, since the other two suits, spades and clubs, are black.

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42.

If the odds in favour of an event are 3:4, what is the probability of the event occurring ?

A.3/7
B.4/7
C.3/4
D.7/3

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

The probability is 3/7.

Odds in favour of 3 : 4 mean 3 favourable outcomes for every 4 unfavourable ones, 7 in all. So the probability of the event is 3/(3 + 4) = 3/7.

4/7 is the probability of the event not happening, 3/4 is the odds written as a fraction, and 7/3 is greater than 1, which no probability can be.

In general, odds of a : b in favour give a probability of a/(a + b).

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43.

Which of the following is a characteristic of a binomial distribution ?

A.The events are dependent
B.The outcomes are continuous
C.There are two outcomes, success or failure
D.The Mean is equal to the Mode

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: C

A binomial distribution has two outcomes, success or failure, on each trial.

The binomial distribution counts successes in a fixed number n of independent trials, each with only two outcomes and the same probability of success p.

The trials are independent, not dependent, and the count of successes is discrete, not continuous. The mean (np) need not equal the mode, so that is not a defining feature.

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44.

Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty ?

A.pq+(1-p)(1-q)
B.(1-q)p
C.(1-p)q
D.pq

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

The probability is \(pq + (1-p)(1-q)\).

A computer is declared faulty in two ways:

it is faulty (probability p) and the test is correct (probability q): \(pq\);
it is good (probability 1 - p) and the test is wrong (probability 1 - q): \((1-p)(1-q)\).

Adding the two disjoint cases gives the answer. \(pq\) alone counts only the correctly detected faults, while \((1-q)p\) and \((1-p)q\) are the cases where a faulty machine passes or a good machine is correctly passed.

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45.

X is a uniformly distributed random variable that takes value between 0 and 1.The value of E(X³) will be

A.0
B.1/8
C.1/4
D.1/2

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: C

\(E(X^{3}) = \tfrac{1}{4}\).

For X uniform on (0, 1) the density is 1, so

\[ E(X^{3}) = \int_{0}^{1} x^{3}\,dx = \frac{1}{4} \]

1/2 is \(E(X)\) and 1/8 is \(E(X)^{3}\), the cube of the mean, which is not the mean of the cube. Zero would need X to be symmetric about 0.

In general, for X uniform on (0, 1), the mean of the k-th power is 1/(k + 1), so the second moment is 1/3 and the third 1/4.

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46.

What is the probability of observing exactly k successes in n trials in a binomial distribution with parameter p ?

A.\(p(1 - p)^{k}\)
B.\(p^{k}(1 - p)^{n-k}\)
C.\({}^{n}C_{k}\, p^{n}(1 - p)^{k}\)
D.\({}^{n}C_{k}\, p^{k}(1 - p)^{n-k}\)

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: D

The probability is \({}^{n}C_{k}\, p^{k}(1-p)^{n-k}\).

Any particular sequence of k successes and n - k failures has probability \(p^{k}(1-p)^{n-k}\), and there are \({}^{n}C_{k}\) ways of choosing which k of the n trials are the successes.

Without the \({}^{n}C_{k}\) factor you get the probability of one particular sequence only. \(p(1-p)^{k}\) and \({}^{n}C_{k}\, p^{n}(1-p)^{k}\) put the exponents in the wrong places.

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47.

Conditional Probability: In a class, 60% of the students are girls, and 70% of the students like math. If 80% of the girls like math, what is the probability that a randomly selected math-loving student is a girl ?

A.0.68
B.0.25
C.0.50
D.0.42

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

The probability is about 0.68.

By Bayes' theorem, with G = girl and M = likes math:

\[ \begin{aligned} P(G \mid M) &= \frac{P(M \mid G)\,P(G)}{P(M)} \\ &= \frac{0.8 \times 0.6}{0.7} = \frac{0.48}{0.7} \approx 0.686 \end{aligned} \]

0.68 is the closest option. 0.48 would be the joint probability of a girl who likes math, which is not offered, and 0.25, 0.50 and 0.42 do not follow from the data.

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48.

In a group of 100 students, 60 play cricket, 50 play football, and 30 play both. If a student is selected at random, what is the probability that the student plays cricket given that the student plays football ?

A.0.15
B.0.30
C.0.60
D.0.18

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: C

The probability is 0.60.

Given that the student plays football, we look only at the 50 football players, of whom 30 also play cricket:

\[ P(C \mid F) = \frac{30}{50} = 0.60 \]

0.30 is the unconditional probability of playing both (30 out of 100), and 0.15 and 0.18 do not follow from the figures.

The general rule is P(C | F) = P(C and F) / P(F) = 0.30 / 0.50, which gives the same 0.60.

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49.

Find the variance in uniform distribution if X is uniformly distributed over (5, 30)?

A.85.052
B.52.083
C.1/5
D.17.5

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: B

The variance is 52.083.

For a uniform distribution on (a, b) the variance is \((b-a)^{2}/12\):

\[ \frac{(30 - 5)^{2}}{12} = \frac{625}{12} \approx 52.083 \]

17.5 is the mean, \((5 + 30)/2\), not the variance, and 1/5 and 85.052 do not come from the formula.

For any uniform distribution the mean is the midpoint (a + b)/2 and the variance depends only on the width b - a, so shifting the interval does not change the variance.

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50.

Compute the Median for the exponential distribution with parameter λ :

A.log(3)/λ
B.log(2)/2λ
C.2log(2)/λ
D.log(2)/λ

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: D

The median is \(\log(2)/\lambda\).

For an exponential distribution with rate \(\lambda\), \(P(X \le x) = 1 - e^{-\lambda x}\). The median m makes this one half:

\[ \begin{aligned} 1 - e^{-\lambda m} &= \tfrac{1}{2} \\ m &= \frac{\ln 2}{\lambda} \end{aligned} \]

Here log means the natural logarithm. \(\log(2)/2\lambda\) and \(2\log(2)/\lambda\) are off by a factor of two, and \(\log(3)/\lambda\) would be the point with probability 2/3 below it.

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