Probability & Statistics Ch.2 Section 3 Question 15

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Suppose that a basketball player sinks a basket from a certain position on the court with probability 0.35.

a) What is the probability that the player sinks three baskets in 10 independent throws? b) What is the probability that the player throws 10 times before obtaining the first basket? c) What is the probability that the player throws 10 times before obtaining two baskets?

Solution

(a) What is the probability that the player sinks three baskets in 10 independent throws?

This uses the Binomial distribution (independent trials, fixed number of trials).

P(Y=y) = \binom{n}{y}\theta^y(1-\theta)^{n-y}

With n = 10, θ = 0.35, y = 3:

P(Y=3) = \binom{10}{3}(0.35)^3(0.65)^7

= 120 \times 0.042875 \times 0.0490224

P(Y=3) \approx 0.2522

(b) What is the probability that the player throws 10 times before obtaining the first basket?

This uses the Geometric distribution (number of failures before first success).

p_X(x) = (1-\theta)^x\theta

We want 9 failures before a success (10 throws total, with the 10th being the first success):

P(X=9) = (0.65)^9(0.35)

P(X=9) \approx 0.00721

(c) What is the probability that the player throws 10 times before obtaining two baskets?

This uses the Negative-Binomial distribution (trials stop after a certain number of successes).

p_K(k) = \binom{r-1+k}{k}\theta^r(1-\theta)^k

Where:

r = 2 (desired number of successes)
k = 10 - 2 = 8 (number of failures)
θ = 0.35

P(K=8) = \binom{9}{8}(0.35)^2(0.65)^8

= 9 \times 0.1225 \times 0.03186

P(K=8) \approx 0.0351