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 are given the following:
- Total number of students: 9
- Number of minors (not of legal age): 4
- Number of legal age students: 5
- The witness randomly checks the IDs of 5 students
- We are to find the probability that exactly 2 minors are among those 5 students whose IDs are checked.
Step 1: Understanding the problem
We are sampling 5 students at random from a group of 9 students, where 4 are minors and 5 are not. We want the probability that exactly 2 minors are selected (which means the remaining 3 selected must be adults).
This is a hypergeometric probability problem because we are drawing without replacement from a finite population.
Step 2: Hypergeometric Probability Formula
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 population)
- $K = 4$ (number of minors)
- $n = 5$ (number of students selected)
- $k = 2$ (number of minors desired in the selection)
$$
P(\text{exactly 2 minors}) = \frac{\binom{4}{2} \cdot \binom{5}{3}}{\binom{9}{5}}
$$
Step 3: Calculate
- $\binom{4}{2} = 6$
- $\binom{5}{3} = 10$
- $\binom{9}{5} = 126$
$$
P = \frac{6 \cdot 10}{126} = \frac{60}{126} = \frac{10}{21}
$$
✅ Final Answer: $\boxed{\frac{10}{21}}$ or approximately 0.476
Explanation (300 words):
This problem is an example of a hypergeometric distribution, which is used when we are randomly selecting items (or people) from a group without replacement. In this case, the “items” are students, and the characteristic of interest is whether a student is a minor or not.
We know that among the 9 total students, 4 are minors (not of legal age) and 5 are of legal age. The witness checks the IDs of 5 students at random. We are asked to find the probability that exactly 2 of the 5 students are minors. This means the remaining 3 students must be of legal age.
Using the hypergeometric formula, we calculate the number of ways to choose 2 minors from the 4 available ($\binom{4}{2} = 6$) and multiply it by the number of ways to choose 3 adults from the 5 available ($\binom{5}{3} = 10$). This gives us $6 \times 10 = 60$ favorable combinations.
Next, we calculate the total number of ways to choose any 5 students from the 9 total, which is $\binom{9}{5} = 126$.
The probability is then the ratio of favorable outcomes to total outcomes: $\frac{60}{126} = \frac{10}{21}$.
This result shows that there is about a 47.6% chance that the witness checks exactly two minors among the five students. Understanding and applying hypergeometric distribution is essential in real-life sampling situations where selection is without replacement—such as quality control, audits, or random checks like this one.