a. To construct the probability distribution, we assign probabilities to the three possible outcomes: missing both shots (53%), making one shot (31%), and making both shots (16%). We add up these probabilities to ensure they sum to 1.
Probability of missing both shots: 53% = 0.53
Probability of making one shot: 31% = 0.31
Probability of making both shots: 16% = 0.16
Therefore, the probability distribution is:
P(X = 0) = 0.53
P(X = 1) = 0.31
P(X = 2) = 0.16
b. To find the probability that he makes no more than one of the shots, we need to sum the probabilities of making 0 shots and making 1 shot.
P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.53 + 0.31 = 0.84
Therefore, the probability that he makes no more than one of the shots is 0.84.
c. To find the probability that he makes at least one of the shots, we need to subtract the probability of making 0 shots from 1.
P(X ≥ 1) = 1 - P(X = 0) = 1 - 0.53 = 0.47
Therefore, the probability that he makes at least one of the shots is 0.47.
A basketball player is fouled while attempting to make a basket and receives two free throws. The opposing coach believes there is a 53% chance that the player will miss both shots, a 31% chance that he will make one of the shots, and a 16% chance that he will make both shots.
a. Construct the appropriate probability distribution. (Round your answers to 2 decimal places.)
b. What is the probability that he makes no more than one of the shots? (Round your answer to 2 decimal places.)
c. What is the probability that he makes at least one of the shots? (Round your answer to 2 decimal places.)
1 answer
3 answers