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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

Show Tags

09 Nov 2007, 17:19

4

This post received KUDOS

32

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

47% (01:52) correct
53% (01:00) wrong based on 1061 sessions

HideShow timer Statistics

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?

My answer: There are total of 36 possibilities. You can construct a 6x6 table on paper or mentally to sort of visualize the number of possibilities. There are only 2 cases: 5:3 or 3:5 that one of the cards is a 5 and the sum of 2 cards is 8. Thus, the prob. is 2/36 = 1/18

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?

2/5 3/5 1/18 1/3 1/6

tnguyen707 wrote:

My answer: There are total of 36 possibilities. You can construct a 6x6 table on paper or mentally to sort of visualize the number of possibilities. There are only 2 cases: 5:3 or 3:5 that one of the cards is a 5 and the sum of 2 cards is 8. Thus, the prob. is 2/36 = 1/18

since it is said that the sum is 8, total possibilities = (2, 6) (3, 5) (4,4) (5, 3)(6,2)

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?

2/5 3/5 1/18 1/3 1/6

4,4 6,2 2,6 5,3 3,5

P=2/5
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

1/18 would be the probability of getting an 8 with a 5-3 combination. The question has already narrowed down to the fact that the sum of the 2 cards pulled = 8. All we have to do is to look for all the possible combinations of 8, which happen to be 5. Of these, only 2 have a 5 in them. Therefore, the Prob. = 2/5

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

Show Tags

05 Jun 2012, 01:40

ggarr wrote:

Quote:

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?

2/5 3/5 1/18 1/3 1/6

My soln:

two possibilities:

5,3 and 3, 5

1/6*1/6*2= 1/18

some answers say 2/5 , but since the cards can be replaced , I think 2/5 may not be the correct answer .

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?

2/5 3/5 1/18 1/3 1/6

My soln:

two possibilities:

5,3 and 3, 5

1/6*1/6*2= 1/18

some answers say 2/5 , but since the cards can be replaced , I think 2/5 may not be the correct answer .

can anyone shed some light please ? Thank you

What combinations of two cards are possible to total 8? (first card, second card): (6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened. Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).

From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

Show Tags

20 Aug 2013, 22:09

I'm curious why everyone is counting 4-4 as only one possibility. You could select the first four and then the second four, just like you could have the separate scenario of selecting the second four and then the first four. They are separate cards, aren't they? These two scenarios are just as plausible as selecting 2-6 and 6-2 as different outcomes. What's wrong with my logic?

I'm curious why everyone is counting 4-4 as only one possibility. You could select the first four and then the second four, just like you could have the separate scenario of selecting the second four and then the first four. They are separate cards, aren't they? These two scenarios are just as plausible as selecting 2-6 and 6-2 as different outcomes. What's wrong with my logic?

(4, 4) is one case: first draw = 4 and second draw = 4.

While (2, 6) and (6, 2) are 2 different cases: First draw = 2 and second draw = 6; First draw = 6 and second draw = 2.

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

Show Tags

26 Aug 2014, 05:20

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

Show Tags

26 Aug 2014, 05:32

beckee529 wrote:

i think A) 2/5

possibilities are either (5,3) or (3,5)

picking a five first + picking a five second 1* 1/5 + 1/5 * 1 = 2/5

There is something wrong in this method as there are 6 balls in the bowl and after drawing one ball the ball is put back into the bowl .. its a replacement type of problem . So it will not be 1 *1/5 + 1/5*1

Instead it will be 1*1/6 + 1/6*1 (Replacement type ).

Correct Noob Method :

Possible outcomes:

2-6 3-5 4-4 5-3 6-2 Therefore, total number of outcomes such that the sum is 8 = 5.

Number of outcome such that atleast one card in the sum is 5 = 2 (3-5,5-3).

Re: Six cards numbered from 1 to 6 are placed in an empty bowl. [#permalink]

Show Tags

27 Apr 2015, 17:06

2

This post was BOOKMARKED

ggarr wrote:

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. 2/5 B. 3/5 C. 1/18 D. 1/3 E. 1/6

This is a conditional probability question. The formula for conditional probability says something like -

P (B/A) = P (A and B) / P(A) [Read as Prob. of event B given event A has already occured]

A: event that sum of the numbers on the 2 cards is 8 B : event that one of the 2 cards is numbered 5

Whats P(A and B) ? It means sum is 8 and one of the cards is numbered 5 Well, that's only 2 cases - (5,3) (3,5) n(favorable outcomes) = 2 n(total outcomes) = 6C2 = 15 P(A and B) = 2/15

Substitute in formula P(B/A) = Prob. that one of the 2 cards is 5 given their sum is 8 = (2/15)/(1/3) = 2/5 Ans.

Best Schools for Young MBA Applicants Deciding when to start applying to business school can be a challenge. Salary increases dramatically after an MBA, but schools tend to prefer...

Marty Cagan is founding partner of the Silicon Valley Product Group, a consulting firm that helps companies with their product strategy. Prior to that he held product roles at...