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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

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

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: 3 possible combinations or outcomes [(fruit,salad) - (fruit,soup) - (salad,soup)]
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

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