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 problem, let’s break down the information and apply a hypergeometric probability distribution, which is used when sampling without replacement.
Given:
- Total students = 9
- Minors (underage) = 4
- Legal age students = 9 – 4 = 5
- Witness checks IDs of 5 students, chosen randomly
- We want to find the probability that exactly 2 of the 5 students checked are minors
Approach: Hypergeometric Distribution
The hypergeometric probability formula is: P(X=k)=(Kk)(N−Kn−k)(Nn)P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}P(X=k)=(nN)(kK)(n−kN−K)
Where:
- N=9N = 9N=9 (total population)
- K=4K = 4K=4 (total minors)
- n=5n = 5n=5 (number of students checked)
- k=2k = 2k=2 (number of minors among those checked)
Plug in the values:
P(X=2)=(42)(53)(95)P(X = 2) = \frac{\binom{4}{2} \binom{5}{3}}{\binom{9}{5}}P(X=2)=(59)(24)(35)
- (42)=6\binom{4}{2} = 6(24)=6 (ways to choose 2 minors out of 4)
- (53)=10\binom{5}{3} = 10(35)=10 (ways to choose 3 legal age students out of 5)
- (95)=126\binom{9}{5} = 126(59)=126 (total ways to choose 5 students from 9)
P(X=2)=6×10126=60126=1021P(X = 2) = \frac{6 \times 10}{126} = \frac{60}{126} = \frac{10}{21}P(X=2)=1266×10=12660=2110
Final Answer:
1021≈0.476\boxed{\frac{10}{21}} \approx \boxed{0.476}2110≈0.476
300-Word Explanation:
This problem involves finding the probability that exactly 2 minors are found when randomly checking the ID of 5 out of 9 students, where only 4 students are underage. Since the IDs are being checked randomly without replacement, the hypergeometric distribution is the appropriate model to use.
The hypergeometric distribution calculates the likelihood of obtaining k successes (in this case, minors) from a sample of size n, drawn from a larger population N that contains K total successes. The formula involves three parts:
- Choosing the minors: There are 4 minors, and we want exactly 2 of them in our sample of 5. So, (42)=6\binom{4}{2} = 6(24)=6 ways.
- Choosing the legal-age students: We need the remaining 3 students (since 5 − 2 = 3) to be of legal age. There are 5 such students available, and (53)=10\binom{5}{3} = 10(35)=10 ways to choose them.
- Total combinations: The total number of ways to choose any 5 students out of 9 is (95)=126\binom{9}{5} = 126(59)=126.
Thus, the probability is calculated as: Favorable outcomesTotal possible outcomes=6×10126=60126=1021\frac{\text{Favorable outcomes}}{\text{Total possible outcomes}} = \frac{6 \times 10}{126} = \frac{60}{126} = \frac{10}{21}Total possible outcomesFavorable outcomes=1266×10=12660=2110
This result means there’s approximately a 47.6% chance that the witness randomly checks IDs of exactly two minors when checking five students. This scenario shows how probability can help predict outcomes in random samples, especially when categorizing a fixed number of “successes” within a group.