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

 Let the Probability of success be that the player get the basket in a shot which is given as 0.7 (70%).

X is the number of shots which is given that player takes four independent free throws.

Using, Binomials formula:

$P( x = r) = \begin{pmatrix} n \\ r \end{pmatrix}* P^r( 1 - P)^{n - r}$

where P is the Probability of success, n is number of trials.

In our case , n = 4 ; r =2; P = 0.7

P( exactly 2 baskets) = P(x = 2) $= \begin{pmatrix} 4 \\ 2 \end{pmatrix}*(0.7)^2* ( 1 - 0.7)^{4 - 2} =$

= $[ 4!/ 2!(4-2)!] (0.7)^2 ( 1 - 0.7)^{4 - 2} =$

$= (4*3/2)* (0.7)^2* ( 0.3)^2 =$

$= 6* ( 0.49)* (0.09)$

P( x= 2) = 0.265

Answer: Probability that Raymonds gets exactly 2 baskets is 0.265.