What is the probability that a witness will refuses to serve alcoholic beverages to only two minor if she randomly check Id 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 question, we need to find the likelihood that exactly 2 of the 4 minors are among the 5 students whose IDs are checked randomly from a total of 9 students (4 minors + 5 of legal age).
Given:
Step-by-Step Solution:
We are solving a hypergeometric probability problem.
The hypergeometric formula is:
$$
P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}
$$
Where:
- $N =$ total population = 9
- $K =$ total number of “successes” in population (minors) = 4
- $n =$ number of draws (students checked) = 5
- $k =$ desired number of successes (minors checked) = 2
Now plug in the values:
$$
P(\text{exactly 2 minors}) = \frac{\binom{4}{2} \binom{5}{3}}{\binom{9}{5}}
$$
$$
= \frac{6 \cdot 10}{126} = \frac{60}{126} = \frac{10}{21}
$$
✅ Correct Answer: $\boxed{\frac{10}{21}}$
Explanation (300+ words):
This problem involves calculating the probability of selecting a specific combination of minors and legal-aged individuals from a larger group, which is a typical scenario for the hypergeometric distribution. The hypergeometric distribution is used when sampling is without replacement, meaning once a student is selected to have their ID checked, they are not considered again.
In this situation, we are told that there are 9 students total: 4 minors (underage) and 5 of legal age. The witness randomly checks 5 students’ IDs and we are to find the chance that exactly 2 of those 5 students are minors.
This is not a binomial distribution because we are not assuming independent trials or replacement. Instead, we use combinations to find how many favorable outcomes exist over how many total possible outcomes.
Multiplying $\binom{4}{2}$ and $\binom{5}{3}$ gives us the number of favorable outcomes where exactly 2 minors are chosen. Dividing by the total possible outcomes gives us the probability.
After calculating the combinations and simplifying, we find that the probability is $\boxed{\frac{10}{21}}$, which is about 0.476 or 47.6%. This means that if the witness randomly checks IDs from the group, there’s a 47.6% chance that exactly 2 of the checked students will be minors, leading to their refusal of alcoholic beverages.