Question

4. Up until the last game, Sellina has made 75% of her free-throws. If she continues

to make her free-throws at this rate, what is the probability that she will make the

next two free-throws in the next game?

Answer 1

To find the probability that Sellina will make both free-throws in the next game, multiply the probability of making the first free-throw by the probability of making the second free-throw given that she made the first. The probability is 0.6375, or 63.75%.

Step-by-step explanation:

To find the probability that Sellina will make both free-throws in the next game, we need to multiply the probability of making the first free-throw by the probability of making the second free-throw given that she made the first.

P(D|C) = 0.85

P(C) = 0.75

To calculate the probability, multiply these two probabilities together: P(C) * P(D|C) = 0.75 * 0.85 = 0.6375.

Therefore, the probability that Sellina will make both free-throws in the next game is 0.6375, or 63.75%.

Answer 2

0.5625 = 56.25% probability that she will make the next two free-throws in the next game.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either Sellina makes it, or she misses. The probability of making a free throw is independent of any other free throw. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

75% of her free-throws.

This means that
`p = 0.75`

What is the probability that she will make the next two free-throws in the next game?

This is P(X = 2) when n = 2. So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 2) = C_(2,2).(0.75)^(2).(0.25)^(0) = 0.5625`

0.5625 = 56.25% probability that she will make the next two free-throws in the next game.