What is the probability that a witness will refuses to serve alcoholic beverages to only two minor if she randomly check If of five students from among nine students of which four are not of legal age ?
The correct answer and explanation is :
To determine the probability that a waitress will refuse to serve alcoholic beverages to exactly two minors when she randomly checks the IDs of five students out of a group of nine—four of whom are minors—we can apply the hypergeometric distribution. This distribution is appropriate because we are dealing with selections without replacement from a finite population.
Problem Breakdown:
- Total number of students (population size, ( N )): 9
- Number of minors (successes in the population, ( K )): 4
- Number of students checked (sample size, ( n )): 5
- Desired number of minors among those checked (desired successes, ( k )): 2
Hypergeometric Probability Formula:
The probability ( P(X = k) ) of selecting exactly ( k ) minors when checking ( n ) students from a population of ( N ) students containing ( K ) minors is given by:
[
P(X = k) = \frac{\binom{K}{k} \binom{N – K}{n – k}}{\binom{N}{n}}
]
Where ( \binom{a}{b} ) denotes a binomial coefficient, calculated as:
[
\binom{a}{b} = \frac{a!}{b!(a – b)!}
]
Calculating the Probability:
- Total Ways to Choose 5 Students from 9:
[
\binom{9}{5} = \frac{9!}{5!(9 – 5)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126
]
- Ways to Choose 2 Minors from 4:
[
\binom{4}{2} = \frac{4!}{2!(4 – 2)!} = \frac{4 \times 3}{2 \times 1} = 6
]
- Ways to Choose 3 Non-Minors from 5:
[
\binom{5}{3} = \frac{5!}{3!(5 – 3)!} = \frac{5 \times 4}{2 \times 1} = 10
]
- Calculating the Desired Probability:
[
P(X = 2) = \frac{\binom{4}{2} \times \binom{5}{3}}{\binom{9}{5}} = \frac{6 \times 10}{126} = \frac{60}{126} = \frac{10}{21} \approx 0.4762
]
Conclusion:
The probability that the waitress will refuse to serve alcoholic beverages to exactly two minors when she randomly checks the IDs of five students from this group is ( \frac{10}{21} ), which is approximately 0.4762 or 47.62%. This means that, under these conditions, there’s about a 47.62% chance that exactly two of the five students checked will be minors.
