If a player makes an average of 60% of his free throw attempts, what is the probability he will make both shots of a one-and-one (to have an opportunity to shoot a second free throw he must make the first)?
To find the probability of making both shots of a one-and-one in basketball, we can use the concept of conditional probability.
The probability of making the first shot is given as 60%, which means the player has a success rate of 60% on each free throw attempt. Therefore, the probability of making the first shot is 0.6 or 60%.
Since the player must make the first shot to have an opportunity to shoot a second free throw, the probability of making the second shot would depend on the success of the first shot. If the first shot is missed, there is no opportunity for a second shot.
So, to find the probability of making both shots, we need to calculate the conditional probability of making the second shot given that the first shot was made.
Since the player's success rate is 60% on each attempt, the probability of making the second shot after making the first one would also be 60%. Therefore, the probability of making both shots in a one-and-one situation is:
P(both shots made) = P(first shot made) * P(second shot made | first shot made)
= 0.6 * 0.6
= 0.36 or 36%
Hence, the probability that the player will make both shots of a one-and-one is 36%.