Answer:
The average number of points this player will get in 100 one-and-one free throw situations is 70.
Explanation:
For each free throw, there are only two possible outcomes. Either the player makes it, or he does not. The probability of the player making a free throw is independent of other free throws. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:
![E(X) = np](https://img.qammunity.org/2021/formulas/mathematics/college/66n16kmn896qth698tyf6rfu48vhaipkmv.png)
70% free throw percentage.
This means that
![p = 0.7](https://img.qammunity.org/2021/formulas/mathematics/high-school/mgkca71qc5cdp502iri38ylfbb5mvcnnji.png)
What is the average number of points this player will get in 100 one-and-one free throw situations?
This is E(X) when n = 100. So
![E(X) = np = 100*0.7 = 70](https://img.qammunity.org/2021/formulas/mathematics/high-school/n8lpgisl5kumi6nrar0lqcx104xntp2nue.png)
The average number of points this player will get in 100 one-and-one free throw situations is 70.