Question

determine which scenario could be found using $p(a) = \frac{(_5c_2)(_8c_1)}{_{13}c_3}$

• 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:

To determine which scenario matches the probability formula \( P(A) = \frac{_5C_2 \cdot _8C_1}{_{13}C_3} \), we analyze each option:

Step 1: Recall Combinations Formula

The combination formula is \( _nC_r = \frac{n!}{r!(n - r)!} \), used for choosing \( r \) items from \( n \) without order. The numerator \( _5C_2 \cdot _8C_1 \) suggests choosing 2 from 5 and 1 from 8, while the denominator \( _{13}C_3 \) suggests choosing 3 from \( 5 + 8 = 13 \).

Step 2: Analyze Each Option

• Option 1 (lock code): Lock codes involve permutations (order matters), not combinations. Eliminate.
• Option 2 (winners): Choosing winners (first, second, third) is a permutation (order matters), not a combination. Eliminate.
• Option 3 (committee): 5 men, 8 women. Choosing 2 men from 5: \( _5C_2 \); 1 woman from 8: \( _8C_1 \). Total people: \( 5 + 8 = 13 \); choosing 3: \( _{13}C_3 \). Matches the formula.
• Option 4 (marbles): Yellow marbles = 3 (not 5), so \( _5C_2 \) does not apply. Eliminate.

Answer:

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