Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Apr 2015, 13:36

### 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

# Q14: Six cards numbered from 1 to 6 are placed in an empty

Author Message
TAGS:
Manager
Joined: 11 Jan 2007
Posts: 202
Location: Bangkok
Followers: 1

Kudos [?]: 10 [0], given: 0

Q14: Six cards numbered from 1 to 6 are placed in an empty [#permalink]  08 Jun 2007, 19:50
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Q14:
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5 ?

A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3
_________________

cool

Manager
Joined: 14 Mar 2007
Posts: 238
Followers: 1

Kudos [?]: 1 [0], given: 0

ok,let´s try this one...

in order to get an sum 8 from 6 cards numbered 1 to 6 : 1 2 3 4 5 6
with repetition one can have: 2+6 or 3+5 or 4 + 4 or
6+2 or 5+3 or 4+ 4

so the probability that the first card could be a 5 is 1/6
and the probability the second card could be a 5 is 1/6

so 2* 1/6 = 1/3

C????
Director
Joined: 13 Mar 2007
Posts: 545
Schools: MIT Sloan
Followers: 4

Kudos [?]: 28 [0], given: 0

(D)

This is a conditional probability problem.

P(A|B) = P(A and B) / P(B)

P(A and B) = prob of one of the cards is 5 and sum is 8

5,3 or 3,5 , prob = 2/36 = 1/18

P(B) = prob of sum of 8 = 5/36

now required answer = (1/18)/(5/36) = D
Manager
Joined: 14 Mar 2007
Posts: 238
Followers: 1

Kudos [?]: 1 [0], given: 0

yes, it appears that we need the prob of getting an 8 ....

3+5
4+4
2+6
5+3
6+2

5/36...

ok I see so 2*1/6*1/6/5/36
Manager
Joined: 23 Dec 2006
Posts: 136
Followers: 1

Kudos [?]: 11 [0], given: 0

(D)

This is a conditional probability problem.

P(A|B) = P(A and B) / P(B)

P(A and B) = prob of one of the cards is 5 and sum is 8

5,3 or 3,5 , prob = 2/36 = 1/18

P(B) = prob of sum of 8 = 5/36

now required answer = (1/18)/(5/36) = D
Good call. I didn't realize that it had to be conditional and I was stuck there wondering why I kept getting 2/36.
Senior Manager
Joined: 04 Mar 2007
Posts: 442
Followers: 1

Kudos [?]: 16 [0], given: 0

Getting 8 :
2+6
6+2
3+5
5+3
4+4
total 5 options
there are 2 favorable: 3+5 and 5+3
So the prob witll be 2/5

D
VP
Joined: 08 Jun 2005
Posts: 1147
Followers: 6

Kudos [?]: 127 [0], given: 0

since we already know that the sum of the numbers on the cards is 8 (no need to find out that probability - wording may be misleading !) all we need to do is to figure out what is the probability of geting 5 on one of the draws out of all the "sum 8" draws .

2,6
6,2
5,3
3,5

4,4

hence 2/5 the answer is (D).

Similar topics Replies Last post
Similar
Topics:
Six cards numbered from 1 to 6 are placed in an empty bowl. 2 23 Oct 2007, 11:19
Six cards numbered from 1 to 6 are placed in an empty bowl. 1 29 Jul 2007, 21:22
Six cards numbered from 1 to 6 are placed in an empty bowl. 2 28 Jul 2007, 08:51
Six Cards numbered from 1 to 6 are placed in a empty bowl. 3 30 Sep 2006, 07:21
Six cards numbered from 1 to 6 are placed in an empty bowl. 7 22 Apr 2006, 19:36
Display posts from previous: Sort by