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p(b) question

by kat187 » Wed Mar 17, 2010 9:24 pm
I got this one from Gmat prep

it say there are

15 S rocks
20 P rocks
10 G rocks.

randomly distributed if 2 rocks chosen at random and without replacement what is p(b) that both rocks will be S?

I did 15 s rocks of total = 1/3

so I times 1/3 * 1/3 = 1/9 . Gmat doesn't like that answer
they have 7/66
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by [email protected] » Wed Mar 17, 2010 9:32 pm
kat187 wrote:I got this one from Gmat prep

it say there are

15 S rocks
20 P rocks
10 G rocks.

randomly distributed if 2 rocks chosen at random and without replacement what is p(b) that both rocks will be S?

I did 15 s rocks of total = 1/3

so I times 1/3 * 1/3 = 1/9 . Gmat doesn't like that answer
they have 7/66
P(B) = (15/45)*(14/44) = 7/66 [14/44 because you have only 14 S rocks and 44 total rocks left as you ve already picked one S rock the first time around(there is no replacement - you dont put the first S rock back in the bag). )
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by Stuart@KaplanGMAT » Thu Mar 18, 2010 12:07 pm
kat187 wrote:I got this one from Gmat prep

it say there are

15 S rocks
20 P rocks
10 G rocks.

randomly distributed if 2 rocks chosen at random and without replacement what is p(b) that both rocks will be S?
Whenever we attack a probability question with multiple selections, we need to ask: are we working with or without replacement.

Without replacement means that probabilities change after each selection; with replacement means that probabilities remain static.

Here, as abysince1984 notes, we're working without replacement, so we need to change the probability for the second selection.

Probability = (# of desired outcomes)/(total # of possibilities)

1st rock: P(S) = 15/45 = 1/3

one S rock is gone, so we now have 14 S rocks and 44 total rocks.

2nd rock: P(S) = 14/44 = 7/22

We want the probability that BOTH are S rocks, so we MULTIPLY the individual probabilities:

1/3 * 7/22 = 7/66.
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