Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 26 May 2017, 23:51

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

A basket contains 5 apples of which 1 is spoiled and the

Author Message
Manager
Joined: 04 Apr 2007
Posts: 130
Followers: 2

Kudos [?]: 15 [0], given: 0

A basket contains 5 apples of which 1 is spoiled and the [#permalink]

Show Tags

13 Apr 2007, 21:43
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.

A basket contains 5 apples of which 1 is spoiled and the rest are good. If Henry is to select 2 apples at random from the basket simultaneously and at random, what is the probability that the 2 apples selected will include the spoiled apple?

1/5
3/10
2/5
1/2
3/5

Why wouldn't it just be something like this: First choice good apple = 4/5 chance...times... 2nd choice bad apple 1/4 = 4/20 = 1/5? Because the order could be switched and say the first choice is the bad apple so = 1/5 then times the second choice for a good apple (4/4) = 4/20 = 1/5...

Am I supposed to add these two scenarios?
Manager
Joined: 25 Mar 2007
Posts: 82
Followers: 1

Kudos [?]: 9 [0], given: 0

Show Tags

14 Apr 2007, 00:44
Say you have a box: { 4 good apples, 1 bad apple }

Now, in order for you to pick two apples where one of them is spoiled, you have to choose 1 good apple from 4 good and 1 bad apple from 1 bad.

remember the statement 'and' means multiply, so:

we have 4 C 1 * 1 C 1 ways of choosing 1 good and 1 bad = 4 * 1 = 4
The total number of ways to choose two apples out of five is 5 C 2 = 10

PS. 4C1 = [4! / (4-1)! * 1!] = 4 and similarly with 5C2 = [5! / (5-2)!*2!] = 10
14 Apr 2007, 00:44
Display posts from previous: Sort by