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# Patrons at a certain restaurant can select two of three

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Director
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Patrons at a certain restaurant can select two of three [#permalink]

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26 Mar 2007, 03:29
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Patrons at a certain restaurant can select two of three appetizers--fruit, soup, and salad--along with two of three vegetables--carrots, squash, and peas. What is the statistical probability that any patron will select fruit, salad, squash, and peas?
a. 1/12

b. 1/9

c. 1/6

d. 1/3

e. 1/2

Director
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26 Mar 2007, 03:54
(B) for me

Probability of choosing desirable appetizers = 2 / 3! (2 means fruit, salad), i.e. P1;
Probability of choosing desirable vegetables = 2/ 3! (2 means squash, peas), i.e. P2.

Statistical probability = P1 x P2 = 2/6 x 2/6 = 1/9

What's OA
Director
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26 Mar 2007, 07:59
First, let's break the probability down into two parts:
1) Appetizers
2) Vegetables

the desired event is selecting fruit and salad for appetizers AND selecting squash and pease for vegetables.
The word "AND" indicates multiplication because it means the occurance or two events together at the same time. Thus, we multiply the desired event for appetizers by the desired event for vegetables

1) Appetizers:
One desired outcome and three possible outcomes --> P(fuid,salad) = 1/3

2) Vegetables:
3 possible combinations or outcomes [(peas,carrots) - (squash,carrots) - (squash,peas)]
One desired outcome and three possible outcomes --> P(peas,squash) = 1/3

Total probability: P(squash,peas) AND P(fruit,salad) = 1/3 x 1/3 = 1/9
Director
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26 Mar 2007, 19:07
thanks for explanations. OA is 1/9.
Manager
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28 Mar 2007, 17:37
why is this approach wrong?

probablility=probability of squash x probability of peas x probablity of fruit x probability of salad
= 1/3*1/3*1/3*1/3
=1/81
Senior Manager
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28 Mar 2007, 22:03
nitinneha wrote:
why is this approach wrong?

probablility=probability of squash x probability of peas x probablity of fruit x probability of salad
= 1/3*1/3*1/3*1/3
=1/81

You don't multiply each probability separately...

the probability of choosing squash and peas is NOT 1/3 x 1/3 = 1/9 because the patrons can choose "two from a group of three." So you need to calculate the "combination".

So the prob of choosing squash and peas from the vegetables group is
3C2 = 3.

This is also the case for choosing two out of three in the appetizer group.
3C2 = 3

So total possible combinations are 3 x 3 = 9.

Therefore, 1/9. B (as is the OA)

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29 Mar 2007, 00:44
agree (B) 1/9.

2 independent events occur.

P(A+B) = P(A) * p(B) = 1/3*1/3 = 1/9.

using F/T rule P(A) = F/T : F = 1 ,

T=3 :
fruit, soup
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