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

$\ X\sim Bin(n, p)$

$$P(X=x)=\binom{n}{k}p^x(1-p)^{n-x}$$

Given $p=0.7,n=4$

$$P(X=2)=\binom{4}{2}(0.7)^2(1-0.7)^{4-2}=$$

$$={4!\over 2!(4-2)!}(0.21)^2=0.2646\approx0.265$$