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 nC
2 = \(\frac{n*(n - 1)}{2}\)
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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