Statistics and Probability — JEE Main Questions

29 JEE Main practice questions on Statistics and Probability, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. X ~ B(n,p). If P(X=2) = P(X=3) and P(X=2) = 9·P(X=4), find n and p.

  1. n=6, p=1/2
  2. n=8, p=1/3
  3. n=7, p=3/8
  4. n=5, p=2/5

Answer: n=8, p=1/3

Using binomial probability formula: P(X=k)=C(n,k)p^k(1-p)^(n-k). From P(X=2)=P(X=3): C(n,2)/C(n,3)=p/(1-p) => 3/(n-2)=p/(1-p). From P(X=2)=9P(X=4): C(n,2)/C(n,4)=9(p/(1-p))^2 => 12/[(n-2)(n-3)]=9(p/(1-p))^2. Substitute p/(1-p)=3/(n-2) => 12/[(n-2)(n-3)]=81/(n-2)^2 => 12(n-2)=81(n-3) => n=8, then p/(1-p)=3/6=1/2 => p=1/3.

2. Which of the following is an example of a discrete random variable?

  1. Height of a student measured in cm
  2. Number of defective items in a batch of 100
  3. Time taken to complete a race in seconds
  4. Temperature of a city at noon in °C

Answer: Number of defective items in a batch of 100

A discrete random variable takes countable values. The number of defective items in a batch of 100 can be 0,1,2,...,100, which is finite and countable. Height, time, and temperature are continuous as they can take any value in an interval.

3. Two coins: A (fair) and B (p=2/3). A coin is picked at random and tossed 6 times. Given that 5 heads appear, what is the probability that coin B was used?

  1. 0.7375
  2. 0.2625
  3. 0.5000
  4. 0.2818

Answer: 0.7375

By Bayes' theorem: P(B|X=5) = P(X=5|B)*0.5 / [P(X=5|A)*0.5 + P(X=5|B)*0.5]. P(X=5|A)=C(6,5)*(1/2)^6=6/64=0.09375, P(X=5|B)=C(6,5)*(2/3)^5*(1/3)=6*(32/243)*(1/3)=192/729≈0.2634. Then numerator=0.1317, denominator=0.178575, ratio≈0.7375.

4. For a binomial distribution X ~ B(12, 1/4), how many values of k satisfy P(X=k) > P(X=k+1)?

  1. 8
  2. 10
  3. 9
  4. 7

Answer: 9

The ratio P(X=k+1)/P(X=k) = [(12-k)/(k+1)]*(1/3). Probabilities increase when ratio > 1, i.e., k < 2.25. So for k=0,1,2, P(X=k) < P(X=k+1); for k=3,...,11, P(X=k) > P(X=k+1). Thus k=3 to 11 inclusive (9 values) satisfy the inequality.

5. Which of the following is NOT a condition for a binomial experiment?

  1. The number of trials is fixed.
  2. Each trial has exactly two outcomes.
  3. The probability of success changes from trial to trial.
  4. The trials are independent.

Answer: The probability of success changes from trial to trial.

A binomial experiment requires a fixed number of trials, each trial has two outcomes, the probability of success is constant, and trials are independent. Changing probability violates the constant p condition.

6. If X ~ B(5, 1/2), what is P(X = 2 | X ≥ 1)?

  1. 10/30
  2. 10/32
  3. 10/31
  4. 10/33

Answer: 10/31

Using conditional probability formula P(A|B) = P(A∩B)/P(B). Here A: X=2, B: X≥1. Since X=2 implies X≥1, P(A∩B)=P(X=2)=C(5,2)(1/2)^5=10/32. P(X≥1)=1-P(X=0)=1-1/32=31/32. Thus P(X=2|X≥1)=(10/32)/(31/32)=10/31.

7. Box A has a fair coin (P(H)=0.5), Box B has a biased coin (P(H)=0.7). A box is chosen at random, the coin is tossed 5 times, and 4 heads appear. What is the probability that Box A was chosen?

  1. 0.360
  2. 0.156
  3. 0.303
  4. 0.500

Answer: 0.303

P(4 heads|A) = C(5,4)*(0.5)^5 = 5/32 = 0.15625. P(4 heads|B) = C(5,4)*(0.7)^4*0.3 = 0.36015. By Bayes, P(A|4 heads) = (0.15625*0.5) / (0.15625*0.5 + 0.36015*0.5) = 0.15625 / (0.15625+0.36015) ≈ 0.303.

8. For a binomial distribution with n = 50 and p = 0.5, what is the ratio of variance to mean?

  1. 0.75
  2. 0.25
  3. 1
  4. 0.5

Answer: 0.5

For a binomial distribution, mean μ = np and variance σ² = np(1-p). The ratio σ²/μ = (1-p). Here p=0.5, so ratio = 0.5. Computed: μ = 50×0.5 = 25, σ² = 50×0.5×0.5 = 12.5, ratio = 12.5/25 = 0.5.

9. In tossing a fair coin three times, let X be the number of heads. Which of the following correctly describes X?

  1. X is a real-valued function on the sample space
  2. X is the number of tails in three tosses
  3. X is the outcome of the first toss only
  4. X is the sum of outcomes of three tosses

Answer: X is a real-valued function on the sample space

A random variable is defined as a real-valued function whose domain is the sample space. Here, X assigns to each outcome the number of heads, which is a real number. Hence option a is correct.

10. For a discrete random variable X with values x_i and probabilities p_i, the mean μ is given by:

  1. μ = Σ (x_i - μ)² p_i
  2. μ = Σ x_i² p_i
  3. μ = Σ x_i p_i
  4. μ = Σ x_i p_i²

Answer: μ = Σ x_i p_i

The mean (expected value) of a discrete random variable is defined as μ = E(X) = Σ x_i p_i, where each value is weighted by its probability. This is the long-run average of X over many trials.

11. For a discrete random variable X, which condition must a probability distribution satisfy?

  1. Each probability is between -1 and 1, and their sum is 1
  2. Each probability is between 0 and 1, and their sum is 0
  3. Each probability is between 0 and 1, and their sum is 1
  4. Each probability is between 0 and 1, and their product is 1

Answer: Each probability is between 0 and 1, and their sum is 1

By definition, a probability distribution of a discrete random variable satisfies 0 ≤ p_i ≤ 1 for each i and Σ p_i = 1. This ensures the probabilities are valid and cover all possibilities.

12. A biased coin with P(head)=0.3 is tossed 10 times. What is the probability of exactly 4 heads?

  1. C(10,6) (0.3)^4 (0.7)^6
  2. C(10,4) (0.3)^4 (0.7)^4
  3. C(10,4) (0.3)^6 (0.7)^4
  4. C(10,4) (0.3)^4 (0.7)^6

Answer: C(10,4) (0.3)^4 (0.7)^6

For a binomial distribution with n=10, p=0.3, q=0.7, the probability of exactly k=4 heads is P(X=4)=C(10,4) p^4 q^(10-4)=C(10,4) (0.3)^4 (0.7)^6. This follows from the binomial pmf formula.

More Mathematics topics

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