There are 7 ingredients at a salad bar that can be placed on : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 16 Jan 2017, 06:35

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

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

Kudos [?]: 18 [0], given: 0

There are 7 ingredients at a salad bar that can be placed on [#permalink]

### Show Tags

17 Mar 2007, 11:54
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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: 921
Followers: 2

Kudos [?]: 46 [0], given: 0

### Show Tags

17 Mar 2007, 19:03
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: 148
Followers: 1

Kudos [?]: 122 [0], given: 0

### Show Tags

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
Display posts from previous: Sort by