Question

A college basketball coach has 12 players on his team. eight players are receiving scholarships, and four are not. the coach decides to select a starting lineup by selecting five names out of a bowl. the probability that 4 of 5 players selected are on scholarship is _____. record answer to three decimal places

Answer 1

Solution: The given problem is a binomial distribution.

The probability that a team member receives a scholarship is $\frac{8}{12} = 0.67$

Therefore, the given problem follows binomial with n = 5 and p = 0.6667

Now the probability that 4 of 5 players selected are on scholarship is:

$P(x=4) = \binom{5}{4} \times 0.6667^{4} (1-0.6667)^{5-4}$

                  $=5 \times 0.1976 \times 0.3333$

                  $=0.329$

Therefore, the probability that 4 of 5 players selected are on scholarship is 0.329