Question

there are 10 boys and 12 girls in the tennis club. the coach wants to select two players to practice first. which statements are true? check all that apply. there is approximately a 27 percent likelihood that one boy and one girl will be chosen to practice first. there is approximately a 52 percent likelihood that one boy and one girl will be chosen to practice first. there is approximately a 19 percent likelihood that two boys will be chosen to practice first. there is approximately a 19 percent likelihood that two girls will be chosen to practice first. there is approximately a 29 percent likelihood that two girls will be chosen to practice first.

Explanation:

Step1: Calculate total number of players

Total players = 10 boys + 12 girls = 22 players.

Step2: Calculate number of ways to choose 2 players

Using combination formula \( C(n, k) = \frac{n!}{k!(n - k)!} \), where \( n = 22 \), \( k = 2 \).
\( C(22, 2) = \frac{22!}{2!(22 - 2)!} = \frac{22\times21}{2\times1} = 231 \).

Step3: Calculate probability of one boy and one girl

Number of ways to choose 1 boy from 10: \( C(10, 1) = 10 \).
Number of ways to choose 1 girl from 12: \( C(12, 1) = 12 \).
Number of favorable outcomes = \( 10\times12 = 120 \).
Probability = \( \frac{120}{231} \approx 0.52 \) or 52%.

Step4: Calculate probability of two boys

Number of ways to choose 2 boys from 10: \( C(10, 2) = \frac{10!}{2!(10 - 2)!} = \frac{10\times9}{2\times1} = 45 \).
Probability = \( \frac{45}{231} \approx 0.19 \) or 19%.

Step5: Calculate probability of two girls

Number of ways to choose 2 girls from 12: \( C(12, 2) = \frac{12!}{2!(12 - 2)!} = \frac{12\times11}{2\times1} = 66 \).
Probability = \( \frac{66}{231} \approx 0.29 \) or 29%.

Answer:

• There is approximately a 52 percent likelihood that one boy and one girl will be chosen to practice first.
• There is approximately a 19 percent likelihood that two boys will be chosen to practice first.
• There is approximately a 29 percent likelihood that two girls will be chosen to practice first.