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: Recall Combination Formula

The combination formula \( _nC_r=\frac{n!}{r!(n - r)!} \) is used for selecting \( r \) items from \( n \) items without regard to order. The probability formula here is \( P(A)=\frac{(\text{Number of favorable outcomes})}{(\text{Total number of possible outcomes})} \), where favorable outcomes are calculated as a product of combinations and total outcomes is a combination.

Step2: Analyze Each Option

• Option 1 (three - digit lock code): For a lock code, the order of digits matters (it's a permutation problem), but the formula uses combinations (\( _nC_r \)), so this is not correct.
• Option 2 (winners from schools): Choosing first - place, second - place, and third - place winners is a permutation problem (order matters), while the formula uses combinations. So this is not correct.
• Option 3 (committee members): We have 5 men and 8 women. We want to choose 2 men from 5 (which is \( _5C_2 \)) and 1 woman from 8 (which is \( _8C_1 \)). The total number of people is \( 5 + 8=13 \), and we are choosing 3 people in total (which is \( _{13}C_3 \)). The probability formula \( P(A)=\frac{(_5C_2)(_8C_1)}{_{13}C_3} \) matches this scenario.
• Option 4 (marbles): We have 3 yellow marbles, but the formula has \( _5C_2 \) (implying 5 yellow marbles), and 4 red marbles, but the formula has \( _8C_1 \) (implying 8 red marbles), and total marbles would be \( 3 + 4+5 = 12\), but the formula has \( _{13}C_3 \) (implying 13 total items). So this does not match.

Answer:

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