Question

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

Answer 1

Answer:

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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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.