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 find the probability that a witness refuses to serve alcoholic beverages to exactly two minors when she randomly checks the IDs of five students out of nine total students (of which four are minors and five are of legal age), we are dealing with a hypergeometric probability distribution.
🔍 Step-by-step Breakdown:
Given:
- Total number of students = 9
- Number of minors (not of legal age) = 4
- Number of legal age students = 5
- The witness checks IDs of 5 students
- We want to find the probability that exactly 2 of those 5 are minors
🎯 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 number of minors
- n=5n = 5: Number of students selected
- k=2k = 2: Desired number of minors among selected students
✅ Plug in the values:
P(X=2)=(42)⋅(53)(95)P(X = 2) = \frac{{\binom{4}{2} \cdot \binom{5}{3}}}{\binom{9}{5}} (42)=6,(53)=10,(95)=126\binom{4}{2} = 6, \quad \binom{5}{3} = 10, \quad \binom{9}{5} = 126 P(X=2)=6⋅10126=60126=1021P(X = 2) = \frac{6 \cdot 10}{126} = \frac{60}{126} = \frac{10}{21}
✔️ Final Answer:
1021\boxed{\frac{10}{21}}
📘 Explanation (300+ words):
This problem uses combinatorics to determine the probability of selecting a specific number of minors when checking IDs. Since the selections are random and without replacement, the hypergeometric distribution applies. This distribution helps when you’re sampling from a population where items can be divided into two categories (in this case, minors and those of legal age), and you’re interested in how many of a certain category end up in your sample.
In our case:
- The population is 9 students.
- 4 students are minors (unfavorable for serving alcohol), and 5 are of legal age.
- The witness randomly chooses 5 students and checks their IDs.
- The event of interest is when exactly 2 of those 5 are minors.
To calculate this, we ask: In how many ways can we choose 2 minors from 4, and 3 legal-age students from 5?
That gives:
- (42)=6\binom{4}{2} = 6: Ways to choose 2 minors.
- (53)=10\binom{5}{3} = 10: Ways to choose 3 of legal age.
- Multiply them: 6×10=606 \times 10 = 60
Now, how many total ways are there to choose any 5 students out of 9?
- (95)=126\binom{9}{5} = 126
So the probability that exactly 2 out of the 5 selected are minors is: 60126=1021\frac{60}{126} = \frac{10}{21}
This means that if the witness randomly checks 5 IDs, there is a 10 out of 21 chance (approximately 47.6%) that exactly two of those students will be minors, thus refusing to serve alcohol to only those two.