There is approximately a 52 percent likelihood that one boy and one girl will be chosen to practice first.
The selection of players is an illustration of combination (or selection)
The expression that represents combination is represented as:
[tex]^nC_r = \frac{n!}{(n - r)!r!}[/tex]
Where:
So, we have:
[tex]^{22}C_2 = \frac{22!}{(22 - 2)!2!}[/tex]
Evaluate the differences
[tex]^{22}C_2 = \frac{22!}{20!2!}[/tex]
Expand the factorials
[tex]^{22}C_2 = \frac{22 \times 21 \times 20!}{20!2!}[/tex]
[tex]^{22}C_2 = \frac{22 \times 21}2[/tex]
[tex]^{22}C_2 = 11 \times 21[/tex]
[tex]^{22}C_2 = 231[/tex]
To select a boy and a girl, we have:
[tex]n = ^{10}C_1 \times ^{12}C_1[/tex]
This gives
[tex]n = 10 \times 12[/tex]
[tex]n = 120[/tex]
The probability that the first selection is a boy and a girl is:
[tex]Pr = \frac{120}{231}[/tex]
Evaluate
[tex]Pr = 0.519[/tex]
Express as percentage
[tex]Pr = 0.52[/tex]
Approximate
[tex]Pr = 52\%[/tex]
Hence, there is approximately a 52 percent likelihood that one boy and one girl will be chosen to practice first.
Read more about combination at:
https://brainly.com/question/11732255
Answer:
The answers are
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.
Explanation:
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