M37-108

Question

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}

Bunuel · Answer

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

Tanisha2819 · Answer

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

Bunuel · Answer

There are 6C2 = 15 possible outcomes of pairs (without considering order) when choosing two cards out of six:

(1,2)
(1,3)
(1,4)
(1,5)
(1,6)
(2,3)
(2,4)
(2,5)
 (2,6) 
(3,4)
 (3,5) 
(3,6)
(4,5)
(4,6)
(5,6)

Out of these 15 pairs, only two of them, (2,6) and (3,5), sum up to 8. Therefore, the probability is 2/15.

VivekSri · Answer

Hi  Tanisha2819

You are wrong in the first part where you have written this

Total outcomes: 6C2

This is wrong because it means that  we can't have sample set (a,b) and (b,a) simultaneously. See you have found the sample set just you have to multiple it be 2, this will ensure that (a,b) and (b,a) can be taken.  Just an example in 6C2 it will either take (6,2) or (2,6) but not both, but here in this case it is possible so your outcome will be

2*(6C2)

Hope you understand.

DhanyaAbhirami · Answer

Total Outcomes = 6 X 5 (First card can be drawn in 6 ways, second card in 5 ways)
You can't get sum 8 by drawing a 1 or a 4
Favourable Outcomes = 4 X 1 (First draw one among 2,3,5,6 say x, and then draw it's counterpart 8-x)
Total ways =  \frac 4 {30}= \frac 2 {15}

M37-108

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