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If there are g girls and b boys on a team, and two members of the team

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If there are g girls and b boys on a team, and two members of the team  [#permalink]

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New post 10 Feb 2019, 01:37
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

58% (02:14) correct 42% (02:19) wrong based on 12 sessions

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

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Re: If there are g girls and b boys on a team, and two members of the team  [#permalink]

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New post 10 Feb 2019, 01:57
1
SajjadAhmad wrote:
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. \(\frac{b(2g+b-1)}{(b+g)(b+g-1)}\)


Lets donate some values

G = 3, B =2, Total number of ways they can be selected = \(5C_2\) = 10

Probability(at least 1 girl) = \(3C_1 4C_1+ 3C_2\) = 9

So are Target value = 9/10

Denominator of all answer options will be 20

Now only D matches our target value

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

3* (7-1) = 3* 6/20 = 9/10
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Re: If there are g girls and b boys on a team, and two members of the team   [#permalink] 10 Feb 2019, 01:57
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