A basketball player takes 20 free throws in a practice session. Each shot is made independently with probability [math: p=0.7]. Which probability distribution is most appropriate for modelling the number of successful shots?[Fill in the blank]Single choice
A
a. Geometric distribution
B
b. Binomial distribution
C
c. Poisson distribution
D
d. Uniform distribution

Question

A basketball player takes 20 free throws in a practice session. Each shot is made independently with probability [math: p=0.7]. Which probability distribution is most appropriate for modelling the number of successful shots?[Fill in the blank]Single choice

A

a. Geometric distribution

B

b. Binomial distribution

C

c. Poisson distribution

D

d. Uniform distribution

BlackTom AI · Accepted Answer

A binomial model fits because there are a fixed number of independent trials, each with two outcomes and the same success probability.

Relevant knowledge points and concepts:
- Binomial distribution conditions: (1) fixed number of trials n, (2) only two outcomes per trial (success/failure), (3) trials independent, (4) constant probability of success p for each trial. If these hold, the number of successes X ~ Binomial(n, p) with P(X = k) = C(n,k) p^k (1 − p)^(n−k).
- Geometric distribution models the number of trials until the first success (not the count of successes in a fixed number of trials).
- Poisson distribution approximates counts of rare events in a fixed interval (typically when n is large and p is small, so np is moderate).
- Uniform distribution assigns equal probability to each outcome in a specified range — not appropriate for success/failure trials with known p.

Step-by-step reasoning:
1. Identify n: the player takes 20 free throws, so n = 20 (fixed).
2. Identify outcomes: each attempt is either made (success) or missed (failure) — two outcomes.
3. Independence: the problem statement says each shot is made independently.
4. Constant p: each shot has the same success probability p = 0.7.
5. All four binomial conditions are satisfied, so the number of successful shots X follows Binomial(20, 0.7). The probability of exactly k successes is P(X = k) = C(20, k) (0.7)^k (0.3)^(20−k).

Why the correct answer is correct:
- The scenario exactly matches the binomial experiment definition, so the binomial distribution is the appropriate model.

Why other options are incorrect:
- Geometric: models trials until first success, not total successes in 20 tries.
- Poisson: used for rare events with small p and large n; here p = 0.7 is not small and n is fixed at 20.
- Uniform: implies equal probability for all counts, which is not true given the known p.

Conclusion: The correct choice is b. Binomial distribution, which matches the provided answer.

A basketball player takes 20 free throws in a practice session. Each shot is made independently with probability [math: p=0.7]. Which probability distribution is most appropriate for modelling the number of successful shots?[Fill in the blank]Single choice

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