Question

A basketball player makes 70% of his free throws the random numbers below represent 20 trials for a simulation let the numbers 0 to 6 represent a made free throw and let 7 to 9 represent a missed free throw use this simulation to estimate the probability that the player makes both free throws. (5,5) (0,6) (9,6) (7,4) (9,1) (7,3) (0,4) (9,3) (2,9) (1,4) (6,8) (5,3) (1,7) (1,7) (1,8) (3,5) (7,2) (2,1) (0,6) (1,8)

Answer 1

Answer:

40%

Step-by-step explanation:

Given the result of the simulation below:

(5,5) (0,6) (9,6) (7,4) (9,1) (7,3) (0,4) (9,3) (2,9) (1,4)

(6,8) (5,3) (1,7) (1,7) (1,8) (3,5) (7,2) (2,1) (0,6) (1,8)

n(S)=20

• The numbers 0 to 6 represent a made free throw .
• The numbers 7 to 9 represent a missed free throw.
• Therefore, any result containing 7 to 9 does not satisfy our requirement that the player makes both throws.

Let Event A be the event that the player made both throws

The pairs in which the player made both throws are:

(5,5) (0,6) (0,4) (1,4) (5,3)  (3,5)  (2,1) (0,6)

n(A)=8

Therefore:

$P(A)=\frac{n(A)}{n(S)}=\frac{8}{20}=0.4$

The probability that the player made both throws is 40%.