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There are 7 ingredients at a salad bar that can be placed on

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Manager
Joined: 29 Nov 2006
Posts: 83
Followers: 1

Kudos [?]: 3 [0], given: 0

There are 7 ingredients at a salad bar that can be placed on [#permalink]  17 Mar 2007, 11:54
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There are 7 ingredients at a salad bar that can be placed on a salad as topping: alfalfa sprouts, radishes, kidney beans, lima beans, scallions, blue cheese and succotash. If a patron at the salad bar chooses three toppings, what is the likelihood that among those toppings will NOT be alfalfa sprouts, lima beans, or succotash?

(A) 1/35
(B) 2/35
(C) 4/35
(D) 9/35
(E) 31/35
Director
Joined: 28 Dec 2005
Posts: 922
Followers: 2

Kudos [?]: 37 [0], given: 0

I get 4/35.

Basically, the prob. that one of those 3 is not chosen is:

P(one of the other 4)* P(of the other 3 left) * P(other 2 left)

= 4/7*3/6*2/5 = 4/35.
Manager
Joined: 20 Jun 2005
Posts: 151
Followers: 1

Kudos [?]: 23 [0], given: 0

Re: PS - Probability [#permalink]  18 Mar 2007, 18:13
There are 7 ingredients at a salad bar that can be placed on a salad as topping: alfalfa sprouts, radishes, kidney beans, lima beans, scallions, blue cheese and succotash. If a patron at the salad bar chooses three toppings, what is the likelihood that among those toppings will NOT be alfalfa sprouts, lima beans, or succotash?

(A) 1/35
(B) 2/35
(C) 4/35
(D) 9/35
(E) 31/35

answer is (C) : C(4,3)/C(7,3) = 4/35.

C(4,3) = 4 is the selection of 3 ingredients from {radishes, kidney beans, lima beans, blue cheese}.

C(7,3) - total number of combinations.
Re: PS - Probability   [#permalink] 18 Mar 2007, 18:13
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