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I'm wondering why it does not work:

1 - P(no five). Since sum us 8, we have 5 possible pairs (2,6), (3,5), (4,4), (5,3), (6,2).
Selecting no five in first turn = 4/5 and in second = 4/5
1 - (4/5 *4/5)
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I'm wondering why it does not work:

1 - P(no five). Since sum us 8, we have 5 possible pairs (2,6), (3,5), (4,4), (5,3), (6,2).
Selecting no five in first turn = 4/5 and in second = 4/5
1 - (4/5 *4/5)
KarishmaB
­This is conditional probability. You have 5 total possible cases in which sum is 8. This is a pre-condition. It must be satisfied by total cases.
So total cases are 5: (2,6), (3,5), (4,4), (5,3), (6,2).
Favorable Cases are 2: (3,5) and (5,3) (the ones which have a 5)
So probability will be 2/5.

Answer (D)

What happens when you do the following:
Selecting no five in first turn = 4/5. Effectively, you have selected a 2 or 3 or 4 or 6.
and in second = 4/5. Again, effectively, you have selected a 2 or 3 or 4 or 6.

So what did you get overall? You got (2, 2), (2, 3), (2, 4).... (3, 2), (3, 4)......
Now this probability you are subtracting from 1 which doesn't work. The pre-condition of a sum of 8 is not satisfied. You have picked all pairs here which do not have a 5 here, whatever the sum may be.
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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) \(\frac{1}{6}\)

(B) \(\frac{1}{5}\)

(C) \(\frac{1}{3}\)

(D) \(\frac{2}{5}\)

(E) \(\frac{2}{3}\)


Attachment:
QR18.png
­According to the given condition possible outcomes are
2,6
6,2
5,3
3,5
4,4
so out of the 5 possible outcomes only 2 have 5 in them.
so answer is 2/5. Option D
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