Question

A professional basketball player is a poor free-throw shooter. Consider situations in which he shoots a pair of free throws. The probability that he makes the first free throw is 0.51. Given that he makes the first, suppose the probability that he makes the second is 0.59. Given that he misses the first, suppose the probability that he makes the second one is 0.36. What is the probability that he makes one of the two free throws?

Answer 1

Answer:

38.55% probability that he makes one of the two free throws

Step-by-step explanation:

We have these following probabilities:

51% probability of making the first free throw.

100-51 = 49% probability of missing the first free throw.

If he makes the first free throw, a 59% probability of making the second free throw and a 100-59 = 41% probability of missing the second free throw.

If he misses the first free throw, a 36% probability of making the second free throw and a 100-36 = 64% probability of missing the second free throw.

What is the probability that he makes one of the two free throws?

Makes first(0.51) and misses second(0.41)

Misses first(0.49) and makes second(0.36)

So

$P = 0.51*0.41 + 0.49*0.36 = 0.3855$

38.55% probability that he makes one of the two free throws

Answer 2

The probability that he makes one of the two free throws is 38.55%.

Calculation of the probability:

Since 51% probability of making the first free throw. And, the first, suppose the probability that he makes the second is 0.59. Given that he misses the first, suppose the probability that he makes the second one is 0.36.

So, the missing the first free-throw probability is

= 100-51

= 49%

Now

The final probability is

$= 0.51 \times 0.41 + 0.49 \times 0.36$

= 38.55%

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