Answer to Question #243473 in Statistics and Probability for bang

Question

what is the probability that a waitress will refuse to serve alcoholic beverages to only 2 minors if she randomly checks the IDs of 6 students from among 9 students of which 4 are not of legal age?

Answer 1

The objective of the problem is determined below:

From the information given, randomly checks the IDs of 5 students from among 9 students of which 4 are not of legal age. That is, N = 9, m = 4, n = 5 .

The objective of the problem is to find the probability that a waitress will refuse to serve alcoholic beverages to only 2 minors.

The objective of the problem is to obtain the probability that a waitress will refuse to serve alcoholic beverages to only 2 minors by using 5 students from among 9 students of which 4 are not of legal age.

The probability that a waitress will refuse to serve alcoholic beverages to only 2 minors are obtained below:

The required probability is,

$h(x; N=9,n=5,k=4) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}}$

where max{0, n-(N-k)} ≤x≤ min{n,k}

i.e. 0 ≤x ≤4

$P(X=2) = \frac{\binom{4}{2} \binom{9-4}{5-2}}{\binom{9}{5}} \\ = \frac{\binom{4}{2} \binom{5}{3}}{\binom{9}{5}} \\ = \frac{\frac{4!}{2!(4-2)!} \times \frac{5!}{3!(5-3)!}}{\frac{9!}{5!(9-5)!}} \\ = \frac{10}{21} = 0.47619$

The probability that a waitress will refuse to serve alcoholic beverages to only 2 minors is obtained by substitute the value of n, m, x, and N in the hypergeometric distribution formula. It can be expected that about 47.621% of the times waitresses will refuse to serve alcoholic beverages to only 2 minors.

Thus the answer is 0.4762