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GMAT Prep Q - Probability

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GMAT Prep Q - Probability

by tito1545 » Thu Mar 03, 2011 4:37 pm
A step by step approach for this please:

A certain manufacturer for a cake , muffin , and bread mixes has 100 buyers , of whom 50 purchase cake mix , 40 purchase muffin mix , and 20 purchase both cake mix and muffin mix.If a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix or muffin mix ?

1.1/10
2.3/10
3.1/2
4.7/10
5.9/10

Ans B
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by Night reader » Thu Mar 03, 2011 5:01 pm
this is combined set and probability problem --> out of 50 who buy cake mix and 40 who buy muffin mix, 20 in each set will buy the both, hence there are 30 who buy only cake mix and 20 who buy only muffin mix. The rest 100-30-20 will buy either both or none. We know that only 20 buy the both which makes 50+20 and we are left with 30. Finally calculate 30 out of 100 to be 3/10

tito1545 wrote:A step by step approach for this please:

A certain manufacturer for a cake , muffin , and bread mixes has 100 buyers , of whom 50 purchase cake mix , 40 purchase muffin mix , and 20 purchase both cake mix and muffin mix.If a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix or muffin mix ?

1.1/10
2.3/10
3.1/2
4.7/10
5.9/10

Ans B
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by BarryLi » Fri Mar 04, 2011 12:48 am
tito1545 wrote:A certain manufacturer for a cake , muffin , and bread mixes has 100 buyers ,
Let C represent cake mix buyers.
Let M represent muffin mix buyers.
Let B represent bread mix buyers.
Let CM represent a cake mix & muffin mix buyer.
etc.

The key to understanding this problem is visualizing the groups in a Venn diagram such as below:
Image

C + M + B + CM + MB + CB + CMB = 100
tito1545 wrote:of whom 50 purchase cake mix ,
C + CM + CB + CMB = 50

tito1545 wrote:40 purchase muffin mix ,
M + CM + MB + CMB = 40
tito1545 wrote:and 20 purchase both cake mix and muffin mix.
CM + CMB = 20

tito1545 wrote:If a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix or muffin mix ?
The question is asking for a buyer of neither cake mix or muffin mix, in other words a buyer of just bread mix (B).

It is extremely important to identify each of the groups in the image:
Image
1) C + CM + CB + CMB (a cake mix buyer)
2) M + CM + MB + CMB (a muffin mix buyer)
3) CM + CMB (a cake mix and muffin mix buyer)

Putting everything together,
B = 100 - [ (C + CM + CB + CMB) + (M + CM + MB + CMB) - (CM + CMB) ]
B = 100 - [ 50 + 40 - 20 ]
B = 30

P(B) = B/100 = 30/100 = 3/10

Answer: B
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by tito1545 » Fri Mar 04, 2011 8:11 am
many thanks
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by GMATGuruNY » Fri Mar 04, 2011 9:06 am
tito1545 wrote:A step by step approach for this please:

A certain manufacturer for a cake , muffin , and bread mixes has 100 buyers , of whom 50 purchase cake mix , 40 purchase muffin mix , and 20 purchase both cake mix and muffin mix.If a buyer is to be selected at random from the 100 buyers , what is the probability that the buyer selected will be one who purchases neither cake mix or muffin mix ?

1.1/10
2.3/10
3.1/2
4.7/10
5.9/10

Ans B
The big idea with overlapping groups: subtract the overlap.

When we count everyone in the first group (those who bought cake mix) and everyone in the second group (those who bought muffin mix), the number who bought both mixes -- the overlap -- gets counted twice. Thus, the overlap needs to be subtracted from the total so that it doesn't get double-counted.

Total cake mix = 40.
Total bread mix = 50.
Overlap = 20.
Thus, total who bought cake mix and/or bread mix = 40+50-20 = 70.

Thus, 100-70 = 30 who bought neither cake mix nor bread mix.

Neither Mix/Total Buyers = 30/100 = 3/10.

The correct answer is B.
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