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 determine the probability that exactly two minors (underage students) are identified when a witness randomly checks the IDs of 5 students from a group of 9 students, 4 of whom are minors.
Key Information:
- Total students = 9
- Minors (not of legal age) = 4
- Adults (legal age) = 5
- IDs checked = 5 students
- Desired outcome: exactly 2 of the 5 checked are minors
Step-by-step Solution:
This is a hypergeometric probability problem, where we calculate the probability of selecting a certain number of successes (minors) without replacement from a finite population.
The formula for the hypergeometric probability is: P(X=k)=(Kk)⋅(N−Kn−k)(Nn)P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}}
Where:
- N=9N = 9 (total students)
- K=4K = 4 (minors in the group)
- n=5n = 5 (number of students selected)
- k=2k = 2 (number of minors desired)
Now plug into the formula: P(X=2)=(42)⋅(53)(95)P(X = 2) = \frac{{\binom{4}{2} \cdot \binom{5}{3}}}{{\binom{9}{5}}}
Calculate the combinations:
- (42)=6\binom{4}{2} = 6
- (53)=10\binom{5}{3} = 10
- (95)=126\binom{9}{5} = 126
P(X=2)=6⋅10126=60126=1021≈0.4762P(X = 2) = \frac{6 \cdot 10}{126} = \frac{60}{126} = \frac{10}{21} \approx 0.4762
✅ Correct Answer:
1021or approximately0.4762\boxed{\frac{10}{21}} \quad \text{or approximately} \quad \boxed{0.4762}
📘 Explanation (Approx. 300 Words):
This problem deals with selecting a subset from a population without replacement. Since there are only 4 minors in a group of 9 students, we want to find the likelihood that exactly two minors are found when 5 IDs are checked.
This is a classic hypergeometric distribution scenario, where we are drawing from a finite population (students) without replacement. The hypergeometric distribution allows us to model the probability of finding a specific number of “successes” (minors, in this case) in a sample of a given size, when the population has a fixed number of those successes.
In our problem, “success” refers to picking a minor. The formula (Kk)⋅(N−Kn−k)(Nn)\frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} calculates the number of ways to choose kk minors from KK, and n−kn-k adults from the remaining N−KN-K, divided by the total number of ways to choose nn students from NN.
By substituting the correct numbers, we calculate that there are 60 favorable outcomes (choosing 2 minors and 3 adults), and 126 total ways to pick 5 students out of 9. So the probability of checking exactly two minors is 1021\frac{10}{21}, or about 47.62%.
This result helps understand how likely it is for underage drinkers to be identified in a partial ID check — useful in policy and law enforcement settings.