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, we use combinatorics and probability concepts.
🔹 Problem Summary
- There are 9 students in total.
- 4 are minors (not of legal age).
- The witness checks IDs of 5 students at random.
- We want the probability that exactly 2 minors are among those 5 students checked.
✅ Correct Answer:
P=(42)⋅(53)(95)=6⋅10126=60126=1021P = \frac{{\binom{4}{2} \cdot \binom{5}{3}}}{{\binom{9}{5}}} = \frac{6 \cdot 10}{126} = \frac{60}{126} = \frac{10}{21}
🔍 Step-by-Step Explanation (300+ words)
This is a classic hypergeometric probability problem. We are choosing a sample (of 5 students) from a population (of 9 students) without replacement, and we are interested in the probability of getting exactly 2 minors in this sample.
Step 1: Understand the groups
- Total students: 9
- Minors (unqualified to drink): 4
- Adults (legal age): 9 – 4 = 5
- Sample size: 5
- Desired outcome: exactly 2 minors in the sample
Step 2: Use the hypergeometric probability formula:
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 population
- K=4K = 4: total minors (successes)
- n=5n = 5: number selected
- k=2k = 2: number of minors we want
P=(42)⋅(53)(95)P = \frac{\binom{4}{2} \cdot \binom{5}{3}}{\binom{9}{5}}
Step 3: Calculate each combination
- (42)=6\binom{4}{2} = 6: Ways to choose 2 minors out of 4
- (53)=10\binom{5}{3} = 10: Ways to choose 3 adults out of 5
- (95)=126\binom{9}{5} = 126: Total ways to choose 5 students out of 9
P=6⋅10126=60126=1021P = \frac{6 \cdot 10}{126} = \frac{60}{126} = \frac{10}{21}
Step 4: Interpret the result
The probability that the witness checks the ID of exactly 2 minors (out of the 5 randomly selected students) is 10/21, or approximately 47.62%.
This means that under these random conditions, there is nearly a 48% chance that exactly two minors are identified, helping the witness prevent them from being served alcohol.
✅ Final Answer:
1021\boxed{\frac{10}{21}}