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16 May 2022, 08:11
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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. \(1\)
B. \(\frac{2}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{2}{15}\)
A. \(1\)
B. \(\frac{2}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{2}{15}\)
16 May 2022, 08:11
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Official Solution:
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. \(1\)
B. \(\frac{2}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{2}{15}\)
\(P(sum \ of \ 8) = \)
\(=P(5, 3) + P(3, 5) + P(6, 2) + P(2, 6) = \)
\(=(\frac{1}{6}*\frac{1}{5}) + (\frac{1}{6}*\frac{1}{5}) + (\frac{1}{6}*\frac{1}{5}) + (\frac{1}{6}*\frac{1}{5}) = \)
\(=\frac{2}{15}\). 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.
Practice similar questions HERE and HERE.
Answer: E
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. \(1\)
B. \(\frac{2}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{2}{15}\)
\(P(sum \ of \ 8) = \)
\(=P(5, 3) + P(3, 5) + P(6, 2) + P(2, 6) = \)
\(=(\frac{1}{6}*\frac{1}{5}) + (\frac{1}{6}*\frac{1}{5}) + (\frac{1}{6}*\frac{1}{5}) + (\frac{1}{6}*\frac{1}{5}) = \)
\(=\frac{2}{15}\). 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.
Practice similar questions HERE and HERE.
Answer: E
29 Oct 2023, 16:08
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Bunuel
Hello Bunuel
I solved the question in the following way but got the wrong answer. Could you help me understand where i went wrong?
Total outcomes: 6C2
Favourable outcomes: 4C2 (Since there are 4 set of cards to pick from such that the sum of the two cards would be 8 (2,6) (3,5) (5,3) & (6,2). So we can pick any two cards from these 4 available.
I understand we are not picking two 4s since replacement is not allowed but we can consider all the other 4 cases since they are all different cards.
Thank you
