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 determine the probability that a witness will refuse to serve alcoholic beverages to exactly two minors when randomly checking the IDs of five students out of nine—four of whom are minors—we can apply the hypergeometric probability distribution.
Problem Breakdown:
- Total students: 9
- Minors (under legal age): 4
- Non-minors (of legal age): 5
- Students whose IDs are checked: 5
- Desired outcome: Exactly 2 minors among the 5 checked students
Hypergeometric Probability Formula:
The hypergeometric probability ( P(X = k) ) is given by:
[ P(X = k) = \frac{\binom{K}{k} \times \binom{N – K}{n – k}}{\binom{N}{n}} ]
Where:
- ( N ) = Total population size
- ( K ) = Total number of successes in the population
- ( n ) = Number of draws
- ( k ) = Number of observed successes
Applying the Values:
- ( N = 9 ) (total students)
- ( K = 4 ) (minors)
- ( n = 5 ) (students checked)
- ( k = 2 ) (minors desired among the checked)
Plugging these values into the formula:
[ P(X = 2) = \frac{\binom{4}{2} \times \binom{5}{3}}{\binom{9}{5}} ]
Calculating the Combinations:
- ( \binom{4}{2} = \frac{4!}{2!(4 – 2)!} = \frac{4 \times 3}{2 \times 1} = 6 )
- ( \binom{5}{3} = \frac{5!}{3!(5 – 3)!} = \frac{5 \times 4}{2 \times 1} = 10 )
- ( \binom{9}{5} = \frac{9!}{5!(9 – 5)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 )
Calculating the Probability:
[ P(X = 2) = \frac{6 \times 10}{126} = \frac{60}{126} = \frac{5}{21} \approx 0.238 ]
Conclusion:
Therefore, the probability that the witness will refuse to serve alcoholic beverages to exactly two minors when randomly checking the IDs of five students is ( \frac{5}{21} ), or approximately 23.8%.
This problem exemplifies the use of the hypergeometric distribution, which is appropriate when dealing with probabilities without replacement from a finite population. Unlike the binomial distribution, which assumes replacement and independence, the hypergeometric distribution accounts for the changing probabilities as individuals are selected without replacement.