there are 10 boys and 12 girls in the tennis club. the coach wants to select two players to practice first…

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.\nthere is approximately a 27 percent likelihood that one boy and one girl will be chosen to practice first.\nthere is approximately a 52 percent likelihood that one boy and one girl will be chosen to practice first.\nthere is approximately a 19 percent likelihood that two boys will be chosen to practice first.\nthere is approximately a 19 percent likelihood that two girls will be chosen to practice first.\nthere is approximately a 29 percent likelihood that two girls will be chosen to practice first.

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.\nthere is approximately a 27 percent likelihood that one boy and one girl will be chosen to practice first.\nthere is approximately a 52 percent likelihood that one boy and one girl will be chosen to practice first.\nthere is approximately a 19 percent likelihood that two boys will be chosen to practice first.\nthere is approximately a 19 percent likelihood that two girls will be chosen to practice first.\nthere is approximately a 29 percent likelihood that two girls will be chosen to practice first.

Answer

Explanation:

Step1: Calculate total number of players

$10 + 12=22$

Step2: Calculate probability of one - boy and one - girl

The number of ways to choose 2 players out of 22 is $C_{22}^2=\frac{22!}{2!(22 - 2)!}=\frac{22\times21}{2\times1}=231$. The number of ways to choose 1 boy out of 10 and 1 girl out of 12 is $C_{10}^1\times C_{12}^1=10\times12 = 120$. The probability $P_1=\frac{120}{231}\approx0.52$ or 52%.

Step3: Calculate probability of two boys

The number of ways to choose 2 boys out of 10 is $C_{10}^2=\frac{10!}{2!(10 - 2)!}=\frac{10\times9}{2\times1}=45$. The probability $P_2=\frac{45}{231}\approx0.19$ or 19%.

Step4: Calculate probability of two girls

The number of ways to choose 2 girls out of 12 is $C_{12}^2=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66$. The probability $P_3=\frac{66}{231}\approx0.29$ 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.