Question

1. An excellent free throw percentage would be something around 90%. That is, such a basketball player would make 90% of the free throws (foul shots) they took. If the player is given 6 chances to take a free throw shot in a game:

1a. Calculate the probability that this type of player makes all 6 of their free throw shots.

1b. Calculate the probability that this type of player misses all 6 of their free throw shots.

1c. Calculate the probability that this type of player makes at least half of their 6 free throw shots.

2. (A continuation of the previous problem.) Suppose that a player makes 835 out of 956 free throw shots in an 89 game season.

2a. Find the probability that this player takes 6 free throw shots in one game.

2b. Find the probability that this player makes at least 10 free throw shots in one game.

2c. Find the probability that this player makes at least 100 free throw shots in ten games.

2d. Find the probability that this player makes 10 free throw shots in a game if you know that they took 12 free throw shots.

Answer 1

1. Let X be the landom variable dendung L number of successful pree shots x will be a Bromiall variable with parameter n = 6 and þ=906=0.9 X-B (6,0.90) a) Probability that player makes all 6 free throw sheets, P(x=6) P(X=6) = 6 x 0.96 0.166 = 0.5314 1 b) Probability that player misses all 6 free throw shats, P(X=0) P(x=0) = ( x 0.9 x 0.16-0 = 0.000001 c) Probability that player makes at least half free throw shots, P(x>3 P(x-2) = P(x-3) + P(x - 4)+P(x-5); P(x-2) ý ( x 0.9 xo.1 - = 0.9987 U