Question

On a womens basketball team, one player can make 61% of free throws she attempts. A success is making a free throw (in basketball, "making a free throw" means the ball successfully went through the hoop.) The random variable X = the number of free throws made out of the seven attempted. Assume that outcomes of free throw attempts can be considered independent of each other. (a) What is the probability that in a given game, she attempts 7 free throws and makes 3 of them? (4 decimal places) (b) What is the probability that in a given game, she attempts 7 free throws and makes all of them? (4 decimal places) (c) What is the probability that in a given game, she attempts 7 free throws and makes 6 or fewer of them? (4 decimal places) (d) What is the mean (expected value) of number of free throws she makes in 7 attempts? (1 decimal place) (e) Of the choices listed below, which ones indicate that X is a binomial random variable? Select all that apply. X represents the number of failed free throws out of seven attempts. There are seven independent trials. There are two possible outcomes, success (making the free throw) and failure (not making the free throw) The probability of making a free throw is different in each trial. X represents the number of successful free throws out of seven attempts. The probability of making the free throw, 61%, is the same for each trial. If the player fails at the first attempt, she is less likely to successfully make the second and third attempts.

Answer 1

a)

Here, n = 7, p = 0.61, (1 - p) = 0.39 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 3)
P(X = 3) = 7C3 * 0.61^3 * 0.39^4
P(X = 3) = 0.1838

b)

Here, n = 7, p = 0.61, (1 - p) = 0.39 and x = 7
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 7)
P(X = 7) = 7C7 * 0.61^7 * 0.39^0
P(X = 7) = 0.0314

c)

Here, n = 7, p = 0.61, (1 - p) = 0.39 and x = 6
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 6).
P(X <= 6) = (7C0 * 0.61^0 * 0.39^7) + (7C1 * 0.61^1 * 0.39^6) + (7C2 * 0.61^2 * 0.39^5) + (7C3 * 0.61^3 * 0.39^4) + (7C4 * 0.61^4 * 0.39^3) + (7C5 * 0.61^5 * 0.39^2) + (7C6 * 0.61^6 * 0.39^1)
P(X <= 6) = 0.0014 + 0.015 + 0.0705 + 0.1838 + 0.2875 + 0.2698 + 0.1407
P(X <= 6) = 0.9687

d)

mean = np
= 7 * 0.61
= 4.3

e)

X represents the number of successful free throws out of seven attempts.