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 problem, we need to find the probability that exactly two minors are identified when the witness randomly checks the IDs of five students from a group of nine students, where four of the nine are minors (not of legal age).
Step 1: Understand the Problem
- Total students = 9
- Minors = 4 (not of legal age)
- Adults = 9 – 4 = 5
- The witness checks the IDs of 5 students at random
- We want the probability that exactly 2 of the 5 checked students are minors
Step 2: Use the Hypergeometric Distribution
The hypergeometric distribution applies because:
- We’re choosing a sample without replacement
- We’re counting successes (minors) in the sample
The hypergeometric probability formula is:
$$
P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{\binom{N}{n}}
$$
Where:
- $N = 9$ (total students)
- $K = 4$ (total minors)
- $n = 5$ (number of students checked)
- $k = 2$ (desired number of minors in the sample)
Step 3: Plug in the Numbers
$$
P(X = 2) = \frac{\binom{4}{2} \cdot \binom{5}{3}}{\binom{9}{5}}
$$
Calculate:
- $\binom{4}{2} = 6$
- $\binom{5}{3} = 10$
- $\binom{9}{5} = 126$
$$
P(X = 2) = \frac{6 \cdot 10}{126} = \frac{60}{126} = \frac{10}{21} \approx 0.476
$$
Final Answer:
$$
\boxed{\frac{10}{21}} \quad \text{or approximately} \quad \boxed{0.476}
$$
Explanation (300+ words)
This problem is a classic example of the hypergeometric probability distribution, which describes situations where objects are selected at random without replacement from a finite population. Here, the population consists of 9 students, 4 of whom are minors. The goal is to determine the probability that exactly 2 of the 5 students selected for ID checking are minors.
When dealing with such problems, we need to consider both favorable outcomes and total possible outcomes. The total number of ways to choose 5 students from 9 is represented as $\binom{9}{5} = 126$. These are all the equally likely combinations the witness could check.
Now, we look at favorable outcomes: choosing exactly 2 minors from 4, and the remaining 3 students from the 5 adults. The number of ways to choose 2 minors out of 4 is $\binom{4}{2} = 6$, and the number of ways to choose 3 adults out of 5 is $\binom{5}{3} = 10$. So, the number of favorable combinations is $6 \times 10 = 60$.
Thus, the probability is the number of favorable combinations over the total combinations:
$$
\frac{60}{126} = \frac{10}{21} \approx 0.476
$$
This means there’s about a 47.6% chance the witness will identify only two minors if they randomly check the IDs of 5 students. This statistical reasoning is important in legal and regulatory environments where random checks are used to monitor compliance.