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cards--28

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cards--28 [#permalink]

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New post 09 Nov 2008, 21:13
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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

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New post 09 Nov 2008, 21:20
D

First, we must figure out how many ways there are to sum 8 from drawing two of these cards. We only have 2-6, 3-5, 4-4, 5-3, 6-2, or five combinations that will give us a sum of 8 when drawn in the manner instructed. 2 of these 5 options contain the card numbered 5, so 2 of 5, or 2/5.

gorden 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. 1/6

B. 1/5

C. 1/3

D. 2/5

E. 2/3

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New post 09 Nov 2008, 21:30
gorden 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. 1/6

B. 1/5

C. 1/3

D. 2/5

E. 2/3


the probability of one card which is numbered 5 is 1/6, then put back, so the probability of the second time is the same

1/6 + 1/6 = 1/3

C

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Re: cards--28 [#permalink]

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New post 09 Nov 2008, 21:38
Your answer doesn't take into account that the sum of the cards must be 8, and then we must figure the probability that a 5 was drawn as one of the two cards that totalled 8. The question does not ask us to find the probability of drawing the number 5 two times in a row.

Sion wrote:
gorden 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. 1/6

B. 1/5

C. 1/3

D. 2/5

E. 2/3


the probability of one card which is numbered 5 is 1/6, then put back, so the probability of the second time is the same

1/6 + 1/6 = 1/3

C

_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 623 [0], given: 32

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Re: cards--28 [#permalink]

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New post 09 Nov 2008, 21:47
jallenmorris wrote:
Your answer doesn't take into account that the sum of the cards must be 8, and then we must figure the probability that a 5 was drawn as one of the two cards that totalled 8. The question does not ask us to find the probability of drawing the number 5 two times in a row.

Sion wrote:
gorden 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. 1/6

B. 1/5

C. 1/3

D. 2/5

E. 2/3


the probability of one card which is numbered 5 is 1/6, then put back, so the probability of the second time is the same

1/6 + 1/6 = 1/3

C


oops.... you are right, i misunderstood the question

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New post 10 Nov 2008, 01:32
IMO B.
looks a conditional probability question..

6 × 6 = 36 total possible ways
Prob of getting a sum of 8 (namely 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2)= 5/36.
now say if we get a 5 on 1st card and 3 on the 2nd, thats only 1 out of 36 possibilities.(1/36)

therefore, probabilty that one of the cards is 5 given that sum is 8 would be (1/36)/(5/36)=1/5.

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New post 10 Nov 2008, 02:40
prasun84 wrote:
IMO B.
looks a conditional probability question..

6 × 6 = 36 total possible ways
Prob of getting a sum of 8 (namely 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2)= 5/36.
now say if we get a 5 on 1st card and 3 on the 2nd, thats only 1 out of 36 possibilities.(1/36)

therefore, probabilty that one of the cards is 5 given that sum is 8 would be (1/36)/(5/36)=1/5.


first card or second card could be 5. Hence, this will be 2/36 and the required probability = 2/5.

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New post 10 Nov 2008, 02:50
yup..thnx for the correction. :P

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Re: cards--28   [#permalink] 10 Nov 2008, 02:50
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