determine which scenario could be found using $p(a)=\frac{(_{5}c_{2})(_{8}c_{1})}{_{13}c_{3}}$ \nprobability…

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

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

Answer

Explanation:

Step1: Recall combination formula

The combination formula is $C(n,r)=\frac{n!}{r!(n - r)!}$, and the probability formula using combinations is $P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. In the given formula $P(A)=\frac{({5}C{2})({8}C{1})}{{13}C{3}}$, the denominator ${13}C{3}$ represents the total number of ways to choose 3 items from 13 items. The numerator $({5}C{2})({8}C{1})$ represents the number of ways to choose 2 items from 5 items and 1 item from 8 items.

Step2: Analyze each option

  • For the lock - code option: There are 5 even and 5 odd numbers from 0 - 9. The total number of ways to choose 3 numbers for the lock - code is not ${13}C{3}$, so this is incorrect.
  • For the winner - choosing option: Choosing first - place, second - place, and third - place winners is a permutation problem, not a combination problem as represented in the given formula, so this is incorrect.
  • For the committee - member option: There are a total of $5 + 8=13$ people. We want to choose 2 men from 5 men (${5}C{2}$) and 1 woman from 8 women (${8}C{1}$), and the total number of ways to choose 3 committee members from 13 people is ${13}C{3}$. This matches the given formula.
  • For the marble - choosing option: The total number of marbles is $3 + 4+5 = 12$, not 13, so this is incorrect.

Answer:

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