Six cards numbered from 1 to 6 are placed in an empty bowl. : PS Archive
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# Six cards numbered from 1 to 6 are placed in an empty bowl.

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

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22 Apr 2006, 19:36
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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
VP
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22 Apr 2006, 23:02
I am not the professor, but will attempt this.

The combinations where the sum is 8 are

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

Total combination = 6
Out of which the number of time one card is 5 is 2

Hence its 2/5
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Re: PS - Probability for Professor [#permalink]

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23 Apr 2006, 05:51
gmat_crack 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

go with 1/5.

2+6 = 8
3+5 = 8
4+4 = 8
5+3 = 8
6+2 = 8

total combination = 5 no of combinations with 5's = 2. prob = 2/5

Last edited by Professor on 23 Apr 2006, 07:14, edited 1 time in total.
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Re: PS - Probability for Professor [#permalink]

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23 Apr 2006, 05:55
Let me slightly change the question.

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 7, what is the probability that one of the two cards drawn is numbered 5?
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23 Apr 2006, 06:56
Let A be the event for sum of the numbers on the cards is 8

and B be the event for one of the two cards drawn is numbered 5

Then the problem requires P(B/A)

Now P(B/A) = P(AB)/P(A)

P(A) = 5/(6*6) as there are 5 successful combination(2,6),(3,5),(4,4),(5,3),(6,2) of 6*6 selections

P(AB) = 2/(6*6) as as there are 2 successful combination(3,5),(5,3) of 6*6 selections

Hence P(B/A) = 2/5
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Re: PS - Probability for Professor [#permalink]

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24 Apr 2006, 05:17
Professor wrote:
Let me slightly change the question.

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 7, what is the probability that one of the two cards drawn is numbered 5?

1/3 it is Pro.
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24 Apr 2006, 06:26
for one of two card to be 5 other has to be 3

1st > probability of getting 5 and then 3
1/6*1/6 = 1/36

2st > probability of getting 3 and then 5
1/6*1/6 = 1/36

probability either of them 5 = 1st or 2nd => 1/36+1/36 => 2/36 => 1/18
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24 Apr 2006, 06:27
I screwed probability again
24 Apr 2006, 06:27
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