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) question
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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). )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
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Whenever we attack a probability question with multiple selections, we need to ask: are we working with or without replacement.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?
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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