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 Post subject: GMAT Probability
PostPosted: Wed Apr 10, 2013 1:48 am 
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Joined: Tue Apr 13, 2010 8:48 am
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A bag contains 10 red jellybeans and 10 blue jellybeans. If 3 jellybeans are removed one at a time, at random and are not replaced, what is the probability that all 3 jellybeans removed from the bag are blue?

A. 9/100
B. 2/19
C. 1/8
D. 3/20
E. 3/10

(B) Use this link for an instructor video explanation:
http://www.youtube.com/watch?v=uG0wBeJCtLE

To determine the probability that several events all take place, you must first determine the probability that each separate event occurs. In this case, to determine the probability of all 3 jellybeans being blue, you must first determine the separate probability of removing a blue jellybean on each trial.

The first time a jellybean is removed, there are 10 blue jellybeans out of a total of 20 jellybeans in the bag. The probability of removing a blue one is 10/20.

The second time a jellybean is removed, there are only 9 blue jellybeans out of a total of 19 (since we are only concerned with the case where the first was blue), so the probability of getting a blue one is 9/19.

The third time, there are 8 blue jellybeans out of a total of 18 in the bag (by the same reasoning), so the probability is 8/18.

To find the total probability, we must multiply the three probabilities together: 10/20 × 9/19 × 8/18. After reducing the fractions to 1/2 × 9/19 × 4/9, we get 2/19.

In the last step, we reduce numerators and denominators first: the 9 in 9/19 cancels out with the 9 in 4/9, and the 2 in 1/2 cancels out when the 4 in 4/9 is reduced to 2. Now, the multiplication is 1 × 1/19 × 2, or 2/19.

The correct answer is choice (B).
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Don't words "and are not replaced" mean that the probability of removing a blue jellybean on each trial is always 1/2?


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 Post subject: Re: GMAT Probability
PostPosted: Wed Apr 10, 2013 1:49 am 
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Joined: Fri Apr 09, 2010 2:11 pm
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Quote:
Don't words "and are not replaced" mean that the probability of removing a blue jellybean on each trial is always 1/2?
The probability of removing a blue jellybean on each trial would be always 1/2 if we returned that jellybean each time we took it. In that case the number of jellybeans would always be constant: 10 blue out of 20.

In our case we do NOT return a jellybean each time we take it. So in our case the numbers each turn are:
10 blue out of 20 (prob. = 10/20)
9 blue out of 19 (prob. = 9/19)
8 blue out of 18 (prob. = 8/18)


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