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­If there are g girls and b boys on a team, and two members of the team are randomly selected, then what is the probability, in terms of g and b, that at least one selected player is a girl?

A. \(\frac{g(g-1)}{(b+g)(b+g-1)}\)

B. \(\frac{b(b-1)}{(b+g)(b+g-1)}\)

C. \(\frac{2bg}{(b+g)(b+g-1)}\)

D. \(\frac{g(2b+g-1)}{(b+g)(b+g-1)}\)

E. \(\frac{b(2g+b-1)}{(b+g)(b+g-1)}\)­

Solution :

Suppose No of Girls - 3 and No of boys - 4

Scenario 1

When at least one girl and one boy then

3C1 * 4C1/7C2 = 4/7

Scenario 2

When there are 2 girls and zero boys then

3C2/7C2 = 1/7

If we add both scenarios then we will get 5/7. Only option D provides us 5/7 as an answer

IMO Option D

 
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Bunuel
­If there are g girls and b boys on a team, and two members of the team are randomly selected, then what is the probability, in terms of g and b, that at least one selected player is a girl?

A. \(\frac{g(g-1)}{(b+g)(b+g-1)}\)

B. \(\frac{b(b-1)}{(b+g)(b+g-1)}\)

C. \(\frac{2bg}{(b+g)(b+g-1)}\)

D. \(\frac{g(2b+g-1)}{(b+g)(b+g-1)}\)

E. \(\frac{b(2g+b-1)}{(b+g)(b+g-1)}\)­


­

i chose a fun way to solve that problem.

The way to go is: Eliminate all choices, that equals zero in case you have 1 Girl, or 0-1 Boys. I. Addition: are unequal zero in case of no girl.

Boy zero: B, C and E are out.
Girl one: A out.

D is the only robust one.
Answer is therefore D)
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