Question

determine which scenario could be found using $p(a)=\frac{(_{5}c_{2})(_{4}c_{1})}{_{13}c_{3}}$
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 combination formula in probability

The formula $P(A)=\frac{(_{5}C_{2})(_{4}C_{1})}{_{13}C_{3}}$ is a probability formula using combinations. Here, $_{n}C_{r}=\frac{n!}{r!(n - r)!}$. The numerator is the product of two combinations which means we are doing two - separate selection tasks from different subsets and the denominator is the total number of ways to select from the combined set.

Step2: Check each option

• Option 1: For choosing two even and one odd number for a three - digit lock code, we don't have subsets of 5 and 4 and a total of 13 elements in a natural way related to even/odd numbers.
• Option 2: Choosing first - place, second - place, and third - place winners from schools with five and eight competitors respectively is a permutation problem (since order matters for winners), not a combination problem in this form.
• Option 3: If we have a group of five men and eight women, and we want to choose two male and one female committee members, the number of ways to choose 2 men out of 5 is $_{5}C_{2}$, the number of ways to choose 1 woman out of 8 is $_{8}C_{1}$, and the total number of ways to choose 3 members out of $5 + 8=13$ is $_{13}C_{3}$. This matches the given formula.
• Option 4: For choosing two yellow marbles and one red marble from a bag with three yellow, four red, and five green marbles, the number of ways to choose 2 yellow out of 3 is $_{3}C_{2}$, the number of ways to choose 1 red out of 4 is $_{4}C_{1}$, and the total number of ways to choose 3 marbles out of $3 + 4+5 = 12$ is $_{12}C_{3}$, which does not match the denominator of $_{13}C_{3}$ in the given formula.

Answer:

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