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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, 02:37
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A
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Question Stats:

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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, 02:57
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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]

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New post 16 Mar 2019, 13:05
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Plug in values and test.
Say that g=2, b=3
P(at least one selected player is a girl) = 1 - (choose 2 boys)
So 1 - (3/5 * 2/4) = 1 - 6/20 = 14/20, this is our target.

Start with C as usual: 2*3*2/5*4 = 12/20 ... too small (notice that denominators are all the same and numerators increase from A to E)

Try D) 2*(6+2-1)/20 = 14/20, this matches, so it's D.
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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 04 May 2019, 05:49
Can anybody solve this question algebraically and show the answer to be D through steps?

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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 May 2019, 13:55
Can some explain the operations and notation here? I don’t know what the letter C means and I see it in all the explanations.

Also, isn’t the 1-x trick the easiest way to solve? I just used the smart numbers of 6 boys and 4 girls. The probability of at least 1 girl is 1 minus the probability that there are no girls. The chance of no girls is 6/10 * 5/9 = 30/90 or 1/3, and 1 - 1/3 = 2/3.

Then just plug 4 and 6 into our formulas to see what comes out.
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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 Sep 2019, 12:18
Gradus wrote:
Can anybody solve this question algebraically and show the answer to be D through steps?

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This please.
I understand the probability space, I understand the g outside the brackets, but I don't understand the bracket contents in the numerator
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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 Sep 2019, 12:30
KanishkM I think it should be 3c1.2c1 + 3c2 = 9

As, it will be 1G,1B or 2G.
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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 Updated on: 11 Sep 2019, 09:04
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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. \(\frac{b(2g+b-1)}{(b+g)(b+g-1)}\)

Hi all, since the algebraic method is requested, I'll show it, but keep in mind no GMAT question would have that amount of calculation required (even for Q50 questions). Before that allow me to show you the quick method.

Case 1: g girls and 0 boys should give us 100% as the answer. Note that all of the denominators in the choices are the same. After plugging in b = 0 the denominator is \(g*(g-1)\). So we need the numerator to be \(g*(g-1)\) for a 100% result after plugging in b = 0. This leaves us with either A or D.

Case 2: 1 girl and 1 boy should result in 100%, so we can eliminate A as it results in 0%, leaving us with only D as the answer.

Finally, for the algebraic method, we start by noting "prob of at least one girl" is the same as "100% minus the prob of 2 boys". So let's start by finding the probability of selecting 2 boys, which is equal to: (# of 2 boy combinations)/(# of any 2 ppl combinations)

We will use nC2 = \(\frac{n*(n - 1)}{2}\)
The number of combinations in total is (b+g) choose 2, which is \(\frac{(b + g)(b + g - 1)}{2}\).
The number of ways of selecting 2 boys out of b boys is \(\frac{b*(b - 1)}{2}\).

Then the division results in \(\frac{b*(b-1)}{(b + g)(b + g - 1)}\). Finally, we must do 100% minus that to find the answer:

100% - \(\frac{b*(b-1)}{(b + g)(b + g - 1)}\) = \(\frac{(b + g)(b + g - 1) - b*(b-1)}{(b + g)(b + g - 1)}\)

Focusing on the numerator: \((b + g)(b + g - 1) - b*(b-1) = b*(b+g-1) + g*(b + g - 1) - b*(b-1) = b*(b-1) + b*g + g*(b + g - 1) - b*(b - 1) = b*g + g*(b + g - 1) = g*(2b +g - 1)\). Therefore we choose D as that matches the numerator.
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Originally posted by TestPrepUnlimited on 10 Sep 2019, 13:00.
Last edited by TestPrepUnlimited on 11 Sep 2019, 09:04, edited 1 time in total.
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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 Sep 2019, 13:16
KanishkM wrote:
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


Should be
Probability(at least 1 girl) = \(3C_1 2C_1+ 3C_2\) = 9
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Re: If there are g girls and b boys on a team, and two members of the team   [#permalink] 10 Sep 2019, 13:16
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