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.
- n=6, p=1/2
- n=8, p=1/3
- n=7, p=3/8
- 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?
- Height of a student measured in cm
- Number of defective items in a batch of 100
- Time taken to complete a race in seconds
- 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?
- 0.7375
- 0.2625
- 0.5000
- 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)?
- 8
- 10
- 9
- 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?
- The number of trials is fixed.
- Each trial has exactly two outcomes.
- The probability of success changes from trial to trial.
- 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)?
- 10/30
- 10/32
- 10/31
- 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?
- 0.360
- 0.156
- 0.303
- 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?
- 0.75
- 0.25
- 1
- 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?
- X is a real-valued function on the sample space
- X is the number of tails in three tosses
- X is the outcome of the first toss only
- 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:
- μ = Σ (x_i - μ)² p_i
- μ = Σ x_i² p_i
- μ = Σ x_i p_i
- 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?
- Each probability is between -1 and 1, and their sum is 1
- Each probability is between 0 and 1, and their sum is 0
- Each probability is between 0 and 1, and their sum is 1
- 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?
- C(10,6) (0.3)^4 (0.7)^6
- C(10,4) (0.3)^4 (0.7)^4
- C(10,4) (0.3)^6 (0.7)^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.