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by sana.noor » Sun Aug 11, 2013 1:48 am
At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

a) 1/12
b) 5/14
c) 4/9
d) 1/2
e) 2/3

OA is B
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by GMATGuruNY » Sun Aug 11, 2013 2:35 am
At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

A. 1/12

B. 5/14

C. 4/9

D. 1/2

E. 2/3

P(NOT selecting all 3 types of tea) = 1 - P(selecting all 3 types of tea).


Let the 3 types of tea be A, B and C.
For all 3 types of tea to be tasted, one type must be selected EXACTLY TWICE.

P(exactly n times) = P(one way) * (total possible ways).

One way to select A exactly twice: AABC
P(A is selected 1st) = 3/9. (Of the 9 cups, 3 contain A.)
P(A is selected 2nd) = 2/8. (of the 8 remaining cups, 2 contain A.)
P(B is selected 3rd) = 3/7. (Of the 7 remaining cups, 3 contain B.)
P(C is selected 4th) = 3/6. (Of the 6 remaining cups, 3 contain C.)
Since we want all of these events to happen, we multiply the probabilities:
3/9 * 2/8 * 3/7 * 3/6 = 1/56.

Total possible ways to select A exactly twice:
AABC is only ONE WAY for A to be selected exactly twice.
Since arrangement of AABC represents a different way to select A exactly twice, the result above must be multiplied by the number of ways to arrange AABC.
The number of ways to arrange 4 elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways to arrange the identical elements.
The reason: when the identical elements swap positions, the arrangement doesn't change.
Since AABC includes two identical elements, we divide by 2!:
Number of ways to arrange AABC = 4!/2! = 12.
Thus:
Number of ways to select A exactly twice = 1/56 * 12 = 3/14.

BBAC and CCAB:
Since any of the 3 teas could be the one selected exactly twice, the result above must be multiplied by 3:
Thus:
P(selecting all 3 types of tea) = 3/14 * 3 = 9/14.

Thus:
P(not selecting all 3 types of tea) = 1 - 9/14 = 5/14.

The correct answer is B.
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by Brent@GMATPrepNow » Sun Aug 11, 2013 5:39 am
sana.noor wrote:At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

a) 1/12
b) 5/14
c) 4/9
d) 1/2
e) 2/3
Let's say that there are 3 kinds of tea: A, B and C, and there are 3 cups of each tea.

First find P(contestant does not taste tea A)
P(contestant does not taste tea A) = P(1st selection is not tea A AND 2st selection is not tea A AND 3rd selection is not tea A AND 4th selection is not tea A)
= P(1st selection is not tea A) x P(2st selection is not tea A) x P(3rd selection is not tea A) x P(4th selection is not tea A)
= 6/9 x 5/8 x 4/7 x 3/6
= 5/42

Now find P(contestant does not taste tea B)
The steps to find this probability will be the same as steps taken to find the above probability.
So, P(contestant does not taste tea B) = 5/42

Likewise, P(contestant does not taste tea C) = 5/42

So, P(contestant does not taste all of the samples) = 5/42 + 5/42 + 5/42
= [spoiler]5/14[/spoiler]
= B

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by Brent@GMATPrepNow » Sun Aug 11, 2013 5:55 am
sana.noor wrote:At a blind taste competition a contestant is offered 3 cups of each of the 3 samples of tea in a random arrangement of 9 marked cups. If each contestant tastes 4 different cups of tea, what is the probability that a contestant does not taste all of the samples?

a) 1/12
b) 5/14
c) 4/9
d) 1/2
e) 2/3
Here's an approach that uses counting techniques:

Let's say that there are 3 kinds of tea: A, B and C, and there are 3 cups of each tea.

First find P(contestant does not taste tea A)
P(contestant does not taste tea A) = (number of ways to select 4 cups that are NOT tea A)/(total number of ways to select 4 cups of tea)

As always, begin with the numerator
Total number of ways to select 4 cups of tea
There are 9 cups and we want to select 4 of them.
Since the order in which we select the 4 cups does not matter, we can use combinations.
We can select 4 of the 9 cups in 9C4 ways (126 ways)


Number of ways to select 4 cups that are NOT tea A
First throw away the 3 cups of tea A. This leaves us with 6 cups of tea that are NOT tea A.
So, any selection of 4 cups will NOT include tea A.
We can select 4 of the 6 cups in 6C4 ways (15 ways)


So, P(contestant does not taste tea A) = 15/126 = 5/42



Using similar logic, we can see that P(contestant does not taste tea B) = 5/42
And P(contestant does not taste tea C) = 5/42

So, P(contestant does not taste all of the samples) = 5/42 + 5/42 + 5/42
= [spoiler]5/14[/spoiler]
= B

Cheers,
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by mgm » Sun Sep 08, 2013 4:46 pm
Mitch,

I used your approach towards solving this but I ended up with a different solution ,

1 - P(All 3 types chosen) = 1 - (3*3*2*2) / 9C4

= 5/7

Here was my thought process(flawed :( )

Total possible ways = Choose 4 cups out of 9 where order does not matter

Fav' Outcomes = 3ways to choose the 1st cup * 3 ways to choose the 2nd * 2 ways for the 3rd * 2 ways for the 4th. ( AABC)


Any comments would be greatly appreciated... :)