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Bunuel
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There are 6 cards numbered from 1 to 6. They are placed into a box, an 28 Jan 2022, 04:57
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77% (01:53) correct 23% (01:47) wrong based on 241 sessions

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There are 6 cards numbered from 1 to 6. They are placed into a box, and then two cards are drawn without replacement. 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. \(\frac{2}{15}\)


M37-108

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Bunuel
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There are 6 cards numbered from 1 to 6. They are placed into a box, an 05 Feb 2022, 10:00
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Bunuel
There are 6 cards numbered from 1 to 6. They are placed into a box, and then two cards are drawn without replacement. 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. \(\frac{2}{15}\)
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\(P(sum \ of \ 8) = \)

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

\( = \frac{1}{6}*\frac{1}{5} + \frac{1}{6}*\frac{1}{5} + \frac{1}{6}*\frac{1}{5} + \frac{1}{6}*\frac{1}{5} = \)

\( = \frac{2}{15}\).

Here (4, 4) case is not possible since we have only one 4 in the box and we are not putting back cards we draw.

Answer: E.
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There are 6 cards numbered from 1 to 6. They are placed into a box, an 28 Jan 2022, 05:04
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Bunuel
There are 6 cards numbered from 1 to 6. They are placed into a box, and then two cards are drawn without replacement. 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. \(\frac{1}{15}\)
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To understand underlying concept better, try the following similar questions:
https://gmatclub.com/forum/there-are-6- ... 55311.html
https://gmatclub.com/forum/there-are-6- ... 80662.html
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There are 6 cards numbered from 1 to 6. They are placed into a box, an 28 Jan 2022, 09:06
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Bunuel
There are 6 cards numbered from 1 to 6. They are placed into a box, and then two cards are drawn without replacement. 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. \(\frac{1}{15}\)
Show more

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

Note: Since its without replacement, we cannot have the case (4, 4) since the card numbered 4 is present only once.

Total ways in which 2 cards can be picked = 6 x 5 = 30

Thus, probability = 4/30 = 2/15

Bunuel - Shouldn't the answer be 2/15?
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There are 6 cards numbered from 1 to 6. They are placed into a box, an 26 Dec 2025, 01:51
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