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m04 q9

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Re: m04 q9 [#permalink] New post 14 May 2014, 22:39
The combination / way of being exactly 2 person will buy among the 3 are --

The Probability of buying for 1 person (Y) - 30% = 3/10

The Probability of buying for 1 person (Y) - 70% = 7/10


A. X Y Y ; The probability will be - 70% * 30% * 30% = (7/10)*(3/10)*(3/10) = 63/1000

B. Y X Y ; The probability will be - 30% * 70% * 30% = (3/10)*(7/10)*(3/10) = 63/1000

C. Y Y X ; The probability will be - 30% * 30% * 70% = (7/10)*(3/10)*(3/10) = 63/1000

The possibility of the above occurrence will be Either (A) or (B) or (C)

So, the total probability will be = 63/1000 + 63/1000 + 63/1000 = 0.189
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Re: M4 Q9 [#permalink] New post 15 May 2014, 04:17
ra011y wrote:
GMAT TIGER wrote:

you need to multiply 0.063 by 3 because there 3 ways 2 visitors can be chosen;

= 3c2 (0.3)^2 (1-0.3)^(3-2) = 3 (0.3^2) (0.7) = 0.189


I have a doubt.

If the question were to state : "what is the probability that any two will buy a pack of candy?" then would the answer be (0.3 x 0.3 x 0.7) = 0.063?

Am still trying to get my head wrapped around this probability thingy.

Thanks!!


I understand it appears confusing, so I hope my explanation will help.

Imagine Kyle, Andy, and Mark were the three people that went to the mall, and each had a 30% chance of buying candy.

The probability that Kyle and Andy get a pack of candy and Mark does not is (3/10)(3/10)(7/10)
The probability that Kyle and Mark get a pack of candy and Andy does not is (3/10)(3/10)(7/10)
The probability that Mark and Andy get a pack of candy and Kyle does not is (3/10)(3/10)(7/10)

If we wanted to see the probability of one of the above specific scenarios (i.e. Kyle and Andy get candy, and Mark does not), the probability would be simply (3/10)(3/10)(7/10). However, since we don't care WHO gets the candy, only that exactly two people get candy, we have to add all of the above probabilities, because any one of them could occur.

So if the problem differentiated between the three people, and said "What is the probability that person 1 and person 2 will get candy, and person 3 will not?", the answer would be your first instinct, which is (3/10)(3/10)(7/10), because it is specifying two of the objects to be chosen.

Since the original does not specify which objects are to be chosen, only that exactly two will be chosen, then you have to consider how many combinations of those objects can possibly be chosen.

You can generalize this by seeing: Total Probability = (Probability of event) x (Number of possible events).

So if Person 1 and Person 2 get candy and Person 3 does not, there is only one possible event. If any two of the three can get candy, there are 3 possible events. Therefore you would arrive at the answer based on the above equation.
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Re: m04 q9 [#permalink] New post 22 May 2014, 07:09
options available:

CCT or TCC or CTC
(0.3x0.3x0.7)0.3
=0.189

Hence, C
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Hope to clear it this time!!
GMAT 1: 540
Preparing again :(

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Re: m04 q9 [#permalink] New post 23 May 2014, 04:58
Think this way:

3 visitors are named as A, B and C.

Exactly two people buy the candy:
Could be 3 cases:
1) A & B buy but not C : 0.3 * 0.3 * 0.7 = 0.63
2) B & C buy but not A: 0.3 * 0.3 * 0.7 = 0.63
3) A & C buy but not B: 0.3 * 0.3 * 0.7 = 0.63

Answer is 0.189
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Re: m04 q9   [#permalink] 23 May 2014, 04:58
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