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:

Step 1: Calculate total number of players

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

Step 2: Calculate number of ways to choose 2 players

The number of ways to choose 2 players from 22 is given by the combination formula \( C(n, k) = \frac{n!}{k!(n - k)!} \), where \( n = 22 \) and \( k = 2 \).
\[
C(22, 2) = \frac{22!}{2!(22 - 2)!} = \frac{22 \times 21}{2 \times 1} = 231
\]

Step 3: 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 (1 boy and 1 girl) = \( 10 \times 12 = 120 \)
Probability = \( \frac{120}{231} \approx 0.52 \) or 52%.

Step 4: Calculate probability of two boys

Number of ways to choose 2 boys from 10: \( C(10, 2) = \frac{10!}{2!(10 - 2)!} = \frac{10 \times 9}{2 \times 1} = 45 \)
Probability = \( \frac{45}{231} \approx 0.195 \) or approximately 19%.

Step 5: Calculate probability of two girls

Number of ways to choose 2 girls from 12: \( C(12, 2) = \frac{12!}{2!(12 - 2)!} = \frac{12 \times 11}{2 \times 1} = 66 \)
Probability = \( \frac{66}{231} \approx 0.2857 \) or approximately 29%. Wait, but let's check again. Wait, \( \frac{66}{231} = \frac{2}{7} \approx 0.2857 \), which is approximately 29%. But wait, the options: "There is approximately a 19 percent likelihood that two girls will be chosen..." is wrong, and "There is approximately a 29 percent likelihood that two girls will be chosen..." is correct? Wait, no, let's recalculate:

Wait, \( C(12, 2) = 66 \), \( C(22, 2) = 231 \). So \( 66 / 231 = 2/7 ≈ 0.2857 \), which is approximately 29%. So the statement "There is approximately a 29 percent likelihood that two girls will be chosen to practice first" is correct. Wait, but let's check the options again:

Options:

• "There is approximately a 27 percent likelihood that one boy and one girl will be chosen to practice first." → Wrong (we got ~52%)
• "There is approximately a 52 percent likelihood that one boy and one girl will be chosen to practice first." → Correct
• "There is approximately a 19 percent likelihood that two boys will be chosen to practice first." → Correct (45/231 ≈ 0.195)
• "There is approximately a 19 percent likelihood that two girls will be chosen to practice first." → Wrong (66/231 ≈ 0.2857)
• "There is approximately a 29 percent likelihood that two girls will be chosen to practice first." → Correct (66/231 ≈ 0.2857 ≈ 29%)

Wait, but let's recheck the two boys: 45/231 ≈ 0.195, so approximately 19%, correct. Two girls: 66/231 ≈ 0.2857 ≈ 29%, correct. One boy and one girl: 120/231 ≈ 0.52, correct.

So the correct statements are:

• 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.

Answer:

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