Probability

mgmt_gmat 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. 61
B. 51
C. 31
D. 52
E. 32

There are following few cases in vise versa fashion in which the sum is 8

2 and 6
3 and 5
4 and 4

5 in all and only 2 favor

Hence, the required probability is [spoiler]2/5[/spoiler].

Amah!! The choices!!

Ok so Mr. GMAT destroyer didn't give us the answer straight out. Well he already has a 760 on his GMAT so he can afford to toy with us. Conversely, I am here to help. So let me expand on Mr. Destroyer's solution hyperbole to clarify.

Mr. Destroyer is correct there are 3 combinations x 2 that make 8
2 + 6 = 8
3 + 5 = 8
4 + 4 = 8

So considering that the 4 card is the same there are 1/5 + 1/5 posibilities = 2/5 or as the answer choices are written to confuse you right down the end 5:2 or 5 2.

The Answer is "D"

mgmt_gmat 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. 61
B. 51
C. 31
D. 52
E. 32

I have a doubt here (2,6)(6,2)(3,5)(5,3)(4,4) are the combinations I agree, but 4,4 is only counted once whereas it should have been counted twice and the denominator should have been 6 ryte?

I think 2/6 = 1/3 is the answer

ajith wrote:

mgmt_gmat 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. 61
B. 51
C. 31
D. 52
E. 32

I have a doubt here (2,6)(6,2)(3,5)(5,3)(4,4) are the combinations I agree, but 4,4 is only counted once whereas it should have been counted twice and the denominator should have been 6 ryte?

I think 2/6 = 1/3 is the answer

No not really. You are looking at the probability of events occuring.

There is one event, lets call it E1 that consists of you drawing a 3 and then drawing a 5. E1=(3->5)
There is also a DISTINCT event, lets call it E2 of you drawing a 5 and then a 3. E2=(5->3)

However there is only one DISTINCT event of drawing a 4 followed by a 4, E=(4->4) If there was somehow a 4A and 4B cards then there would be 2 events but in this situation there is only 1.

mgmt_gmat 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. 61
B. 51
C. 31
D. 52
E. 32

Could you be generous enough to furnish us with the germane choices to the question, mgmt_gmat?