What is the probability that a witness will refuses to serve alcoholic beverages to only two minor if she randomly check If of five students from among nine students of which four are not of legal age ?
The correct answer and explanation is :
To solve this probability problem, we define:
- 9 students in total.
- 4 are minors (not of legal age).
- 5 students are randomly checked.
- We need the probability that exactly 2 out of the 5 checked students are minors.
This follows a hypergeometric distribution because we are selecting students without replacement. The probability mass function of the hypergeometric distribution is:
[
P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
]
Where:
- ( N = 9 ) (total students)
- ( K = 4 ) (total minors)
- ( n = 5 ) (students checked)
- ( k = 2 ) (minors checked)
Substituting these values:
[
P(X = 2) = \frac{\binom{4}{2} \binom{5}{3}}{\binom{9}{5}}
]
We calculate:
[
\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6
]
[
\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10
]
[
\binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126
]
[
P(X = 2) = \frac{6 \times 10}{126} = \frac{60}{126} = \frac{10}{21} \approx 0.4762
]
Explanation:
This means that if the witness checks the IDs of five randomly selected students, there is a 47.62% probability that exactly two of them will be minors. The hypergeometric distribution is used here because the selections are made without replacement, making each draw dependent on previous selections.
This probability helps in understanding how random checking impacts the likelihood of minors being caught in controlled settings such as bars or clubs.