There are 6 cards numbered from 1 to 6. They are placed into a box, an

Question

There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back and then another card is drawn and put back. What is the probability that the sum of the two cards will be 8?

A. \(\frac{2}{3}\)

B. \(\frac{1}{2}\)

C. \(\frac{5}{12}\)

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

E. \(\frac{5}{36}\)

A.  \frac{2}{3}

B.  \frac{1}{2}

C.  \frac{5}{12}

D.  \frac{2}{5}

E.  \frac{5}{36}

M37-107

Bunuel · Accepted Answer

P(sum \ of \ 8) =

=P(5, 3) + P(3, 5) + P(6, 2) + P(2, 6) + P(4, 4) =

= \frac{1}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6} + \frac{1}{6}*\frac{1}{6} =

= \frac{5}{36} .

Answer: E.

Bunuel · Answer

To understand underlying concept better, try the following similar questions: https://gmatclub.com/forum/there-are-6- ... 80664.html https://gmatclub.com/forum/there-are-6- ... 55311.html

sujoykrdatta · Answer

We need to have the sum of the card numbers equal 8
This can be achieved in the following ways:
(2, 6), (3, 5), (4, 4), (5, 3) and (6, 2) => There are 5 different ways

Note: Since its with replacement, we CAN have the case (4, 4) since the card numbered 4 can be picked twice.

Total ways in which 2 cards can be picked (with replacement) = 6 x 6 = 36

Thus, probability = 5/36

Bunuel  - Shouldn't the answer be 5/36?

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