Question

determine which scenario could be found using ( p(a)=\frac{(_{5}c_{2})(_{8}c_{1})}{_{13}c_{3}} )
probability of choosing two even numbers and one odd number for a three - digit lock code
probability of choosing first - place, second - place, and third - place winners from schools with five and eight
competitors, respectively
probability of choosing two male and one female committee members from a group containing five men and eight
women
probability of choosing two yellow marbles and one red marble from a bag containing three yellow marbles, four
red marbles, and five green marbles

Explanation:

Step1: Analyze the combination formula in the numerator

The formula \(_{n}C_{r}=\frac{n!}{r!(n - r)!}\). In \((_{5}C_{2})(_{8}C_{1})\), \(_{5}C_{2}\) represents choosing 2 items from 5, and \(_{8}C_{1}\) represents choosing 1 item from 8.

Step2: Analyze the combination formula in the denominator

\(_{13}C_{3}=\frac{13!}{3!(13 - 3)!}\) represents choosing 3 items from a total of \(5 + 8=13\) items.

Step3: Check each option

• Option 1: For lock - code (permutation as order matters), but the formula uses combinations. So, incorrect.
• Option 2: Choosing winners (order matters as first, second, third - place), but formula uses combinations. So, incorrect.
• Option 3: Total people \(=5 + 8=13\). Choosing 2 men (\(_{5}C_{2}\)) and 1 woman (\(_{8}C_{1}\)), and total ways of choosing 3 people \(_{13}C_{3}\). Correct.
• Option 4: Total marbles \(=3 + 4+5 = 12

eq13\). So, incorrect.

Answer:

probability of choosing two male and one female committee members from a group containing five men and eight women