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A school debate has 4 girls and 6 boys. A total of 3 of the team members wil chosen to participate in the distract debate. What is the probability that 3 girls and no boys wil be selected

Sagot :

Answer: The fraction 1/30

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Explanation:

4 girls + 6 boys = 10 students total

n = 10 people to pick from

r = 3 selections

Use the nCr combination formula

n C r = (n!)/(r!(n-r)!)

10 C 3 = (10!)/(3!*(10-3)!)

10 C 3 = (10!)/(3!*7!)

10 C 3 = (10*9*8*7!)/(3!*7!)

10 C 3 = (10*9*8)/(3!)

10 C 3 = (10*9*8)/(3*2*1)

10 C 3 = (720)/(6)

10 C 3 = 120

This tells us there are 120 different debate teams possible. Order doesn't matter.

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Of those 120 teams, only 4 of them have all three girls.

Consider the girls with code names A,B,C,D

One team is {A,B,C}

Another is {B,C,D}

Another is {A,B,D}

The last is {A,C,D}

Or you could note that there are 4 ways to not select a particular girl, which leads to the four teams of all girls.

Yet another path you can take is to compute nCr for n = 4 and r = 3. You should get 4 C 3 = 4.

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We found there are 4 ways to have a team of all girls and no boys, out of 120 ways to form a team.

4/120 = (1*4)/(30*4) = 1/30 is the final answer

This is approximately equal to 1/30 = 0.0333 = 3.33%

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