Question

determine which scenario could be found using ( p(a)=\frac{(_{5}c_{2})(_{8}c_{1})}{_{13}c_{3}} ).
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
even numbers and one
odd number for a three -
digit lock code
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

The formula \(P(A)=\frac{(_{5}C_{2})(_{8}C_{1})}{_{13}C_{3}}\) is used for combinations. \(_{n}C_{r}=\frac{n!}{r!(n - r)!}\). Here, we are choosing \(2\) from \(5\) and \(1\) from \(8\), then dividing by choosing \(3\) from \(13\).

Step2: Check each scenario

• First scenario: Choosing first - second - third place (order matters, permutations, not combinations).
• Second scenario: Choosing \(2\) men (\(n = 5,r=2\)) and \(1\) woman (\(n = 8,r = 1\)) from a group of \(5 + 8=13\) people. The total number of ways to choose \(3\) people from \(13\) is \(_{13}C_{3}\). This matches the formula.
• Third scenario: Choosing digits for a lock - code (order matters for lock - code, permutations).
• Fourth scenario: Choosing marbles from a bag with different colors. The total number of marbles is \(3+4 + 5=12

eq13\).

Answer:

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