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

Question

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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Rucha.Shukla · Answer

D

Using combination, calculated (n) and (d) separately then combine and simplify.

Since we are randomly selecting 2 people from g girls and b boys, denominator will be 2C(g+b)

For the (n): Atleast one selected player is a girls means addition of 2 possibilities:

Both girls or 1 girl and 1 boy which is essentially 2Cg + 1Cg*1Cb

Simplified and got D, is it correct ?

Posted from my mobile device

Aryaa03 · Answer

IMO-Option-D 
P(Atleast one girl) = 1 - P(No girl)
Say g = 3, b = 4 and 2 no. are to be randomly selected.
P(No girl) = No. of ways in which two boys can be selected from 4 / Total no. of ways for selection
--> P(NG) = 4C2/7C2 = 6/21 = 2/7
--> P(Atleast 1G) = 1-2/7 = 5/7

Substitute g=3 and b=4 in the answer choices, and check it gives 5/7
Denominator is 42 and is common for all choices
For a) 6/42 = 1/7
For b) 12/42 = 2/7
For c) 24/42= 4/7
For d)30/42 = 5/7
For e) 36/42 = 6/7

Option-D matches

Approach-2: 
P(Atleast one girl) = 1 - P(No girl)
 P(Atleast 1 G) = 1 - \frac{b(b-1)}{((b+g)(b+g-1)} 
 = \frac{(b+g)(b+g-1) - b(b-1)}{ (b+g)(b+g-1)}

Solving above, we get Option-D

GmatTest2026 · Answer

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

Shin0099 · Answer

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)

SkyBulb · Answer

either select GB, BG or GG

so GB and BG are similar and say T = b+g
g/T * b/(T-1) + b/T * g/(T-1) = 2gb/T(T-1)

and GG
g/T * (g-1)/T-1 = (g^2 - g)/ T(T-1)

sum both to get D option

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

Prep Toolkit

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