Can someone help please
1 answer:
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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:
The probability that the player made both throws is 40%.