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:
Let’s carefully analyze the problem step-by-step.
Problem Restatement:
- There are 9 students total.
- Among these, 4 are minors (not of legal age).
- A witness randomly checks IDs of 5 students from the 9.
- We want the probability that the witness refuses to serve alcoholic beverages to exactly 2 minors among those 5 checked.
Key points:
- Refusing service means identifying minors.
- We want exactly 2 minors in the sample of 5.
- Total students = 9.
- Minors = 4.
- Legal age students = 9 – 4 = 5.
Approach:
This is a classic hypergeometric probability problem.
Hypergeometric distribution describes the probability of k successes (minors identified) in n draws (students checked), without replacement, from a finite population of size N that contains K successes.
Parameters:
- Population size, N=9N = 9
- Number of successes in population, K=4K = 4 (minors)
- Number of draws, n=5n = 5
- Number of observed successes, k=2k = 2
Hypergeometric Probability Formula:
P(X=k)=(Kk)×(N−Kn−k)(Nn)P(X = k) = \frac{\binom{K}{k} \times \binom{N-K}{n-k}}{\binom{N}{n}}
Where:
- (Kk)\binom{K}{k} = ways to choose k minors from 4 minors.
- (N−Kn−k)\binom{N-K}{n-k} = ways to choose remaining students who are not minors (legal age) from the 5 legal age students.
- (Nn)\binom{N}{n} = total ways to choose any 5 students from 9 students.
Calculate each term:
- (42)=4!2!×2!=244=6\binom{4}{2} = \frac{4!}{2! \times 2!} = \frac{24}{4} = 6
- (53)=5!3!×2!=1206×2=10\binom{5}{3} = \frac{5!}{3! \times 2!} = \frac{120}{6 \times 2} = 10
- (95)=9!5!×4!=362,880120×24=126\binom{9}{5} = \frac{9!}{5! \times 4!} = \frac{362,880}{120 \times 24} = 126
Substitute into formula:
P(X=2)=6×10126=60126=1021≈0.4762P(X=2) = \frac{6 \times 10}{126} = \frac{60}{126} = \frac{10}{21} \approx 0.4762
Final Answer:
The probability that exactly 2 minors are checked (and thus refused) among the 5 students randomly selected is 1021\boxed{\frac{10}{21}} or approximately 0.4762 (47.62%).
Explanation (300 words):
This problem involves the hypergeometric distribution, which applies when sampling is done without replacement from a finite population that has a mix of two types of elements—in this case, minors (underage students) and legal-age students.
Here, the total number of students is 9, with 4 minors and 5 legal-age students. The witness randomly checks 5 students from these 9, and we want to know the probability that exactly 2 of the 5 checked students are minors.
To solve this, we calculate how many ways the witness can pick exactly 2 minors from the 4 minors available, and at the same time pick the remaining 3 students from the 5 legal-age students. The combination formula (nk)\binom{n}{k} tells us the number of ways to select k elements from n elements without regard to order.
The number of ways to pick exactly 2 minors from 4 is (42)=6\binom{4}{2} = 6. The number of ways to pick the remaining 3 students from the 5 legal-age students is (53)=10\binom{5}{3} = 10. Thus, the number of favorable outcomes (exactly 2 minors among the 5 checked) is 6×10=606 \times 10 = 60.
The total number of ways to choose any 5 students from 9 is (95)=126\binom{9}{5} = 126. Hence, the probability is the ratio of favorable outcomes to total outcomes: P=60126=1021≈0.4762P = \frac{60}{126} = \frac{10}{21} \approx 0.4762
This means that if the witness randomly selects 5 students, there is about a 47.6% chance that exactly 2 of them will be minors who should be refused alcohol. This problem is a straightforward application of hypergeometric probability, which is useful whenever sampling is done without replacement and categories are involved.
If you want, I can also help you understand this concept visually or with other examples!