Question

Raymond is a basketball player who takes four independent free throws with 70% probability of getting a basket on each shot. Let X be the number of baskets Raymond gets. Find the probability that he gets exactly 2 baskets, to 3 decimal places.

Answer 1

We have,

X = the random variable denoting the number of baskets Raymond gets

Since, he takes 4 independent free throws therefore X = 0, 1, 2, 3, 4

The probability of getting a basket on each shot is 70% = 0.7

X ~ bin(n = 4, p = 0.7)

Then the p.m.f. of X is given by,

P(X = x) = $\dbinom{4}{x}(0.7)^x(1-0.7)^{4-x}$ , x = 0(1)4

The probability that Raymond gets exactly 2 baskets

= P(X = 2)

= $\dbinom{4}{2}(0.7)^2(1-0.7)^{4-2}$

= 6 x (0.7)² x (0.3)²

= 0.2646 = 0.265 (rounded to 3 decimal places)

Answer: The probability that Raymond gets exactly 2 baskets is 0.265.