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|Author:||questioner [ Sun Apr 22, 2012 2:11 pm ]|
|Post subject:||GMAT Probability|
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?
(B) Use this link for an instructor video explanation:
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).
Don't words "and are not replaced" mean that the probability of removing a blue jellybean on each trial is alway 1/2?
|Author:||Gennadiy [ Sun Apr 22, 2012 2:18 pm ]|
|Post subject:||Re: t.1, qt.28: probability|
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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