M05-10

Question

There are 6 cards numbered 1 through 6. They are placed in a box, and one card is drawn and returned. Subsequently, another card is drawn and returned. If the sum of the two cards drawn was 8, what is the probability that one of the cards drawn was a 5?

A. 1
B. \(\frac{2}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{1}{3}\)

Show Answer

Official Answer

D

A. 1
B.  \frac{2}{3}  
C.  \frac{1}{2}  
D.  \frac{2}{5}  
E.  \frac{1}{3}

Bunuel · Accepted Answer

Official Solution:

There are 6 cards numbered 1 through 6. They are placed in a box, and one card is drawn and returned. Subsequently, another card is drawn and returned. If the sum of the two cards drawn was 8, what is the probability that one of the cards drawn was a 5?

A. 1
B. \(\frac{2}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{1}{3}\)

Show Answer

Official Answer

D

A. 1
B.  \frac{2}{3}  
C.  \frac{1}{2}  
D.  \frac{2}{5}  
E.  \frac{1}{3}

What are the possible combinations of two cards that result in a total of 8?
 
(first card, second card):
 
(6,2), (2,6), (5,3), (3,5), (4,4). 
 
Thus, there are only 5 possible scenarios where the sum is 8. Note that (6,2) case is different from (2,6) case, and likewise (5,3) case is different from (3,5) case.
 
Out of these five possible cases, only two involve drawing a 5. Therefore,  P = \frac{2}{5} .
 
Practice two other similar questions  HERE  and  HERE .

Answer: D

Bunuel · Answer

This is a conditional probability question.

The question does NOT simply ask about the probability of the sum being 8. It says:   IF  the sum of the two cards was 8, what is the probability that one of the cards drawn was a 5?  So, the question says that IF the two cards are drawn, and the sum is 8, what is the probability that one of the cards drawn was a 5? We can have only 5 possible cases with the sum of 8:
(6,2)
(2,6) 
(5,3) 
(3,5) 
(4,4)
 
From this five cases, only in two we have 5.

munukutlask · Answer

Quick Question: Why do we make the distinction that 5-3 and 3-5 are separate events while we leave 4-4 and the other possibility 4-4 as one single event? Why is E not the right answer? 
Wouldn't the cases for a sum of eight be 3-5, 5-3, 4-4, 4-4, 2-6, 6-2 yielding 1/3 as the conditional probability of picking a 5?

Thanks.

Bunuel · Answer

(4, 4) is one case: first draw = 4 and second draw = 4.

While (2, 6) and (6, 2) are 2 different cases:
First draw = 2 and second draw = 6;
First draw = 6 and second draw = 2.

DarkMight · Answer

Dear Bunuel

I tackled the problem quite similar, but as the question is "what's the probability that one of the cards drawn was a 5" I wrote down the 5 Scenarios and counted out of the 10 cards drawn that 2 are 5-cards. So i said 2/10 are 5-cards and thus P=0.20. Where do I have the logic twist?

Bunuel · Answer

We have only 5 possible cases, not 10:
(6,2)
(2,6) 
( 5 ,3) 
(3, 5 ) 
(4,4)

Two cases have 5's, so P = 2/5.

stark026arya · Answer

To get the sum 8 with number 5,we need (5,3) or (3,5).
The probability is (1/6*1/6)+(1/6*1/6)=1/18

Why this approach is wrong.Please explain.

Lampard42 · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I've revised the question and solution, incorporating additional details for improved clarity. I trust this makes it more comprehensible.

laurendast29 · Answer

I made the assumption from "cards numbered 1 through 6" that 4+4 was not a possiblity, as there would only be one 4 card in the deck.

Bunuel · Answer

Have you paid attention to the highlighted part of the question?

There are 6 cards numbered 1 through 6. They are placed in a box, and one card is drawn and returned. Subsequently, another card is drawn and returned. If the sum of the two cards drawn was 8, what is the probability that one of the cards drawn was a 5?

A. 1
B. \(\frac{2}{3}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{5}\)
E. \(\frac{1}{3}\)

Show Answer

Official Answer

D

tanishqgirotra · Answer

why are we considering 2,5 and 5,2 and the other one as different cases here? And, how to identify when to consider a different case and when to not?

Bunuel · Answer

I think your doubt about the cases is addressed in the thread above. As for your general question, check the links below:

22. Probability 
     Theory   
 Math: Probability 
 Probability Made Easy! 
 GMAT Tip of the Week: 99 Problems But Probability Ain’t One 
     Questions   
 Probability and Geometry Problems 
 Advanced Topic: Probability and Combinations Questions With Solutions 
 DS Questions 
 PS Questions

For more:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

s_clw · Answer

I like the solution - it’s helpful.

voluptatemid · Answer

I don’t quite agree with the solution. Can we not take 4,4 as two different counts - making it 2/6 = 1/3?

Bunuel · Answer

No, that is wrong.  Pairs like (6,2) and (2,6) are different because the first card can be 6 and the second 4, or the first card can be 4 and the second 6. Those are two distinct outcomes. However, for (4,4), both cards must be 6, which means the first card is 6 and the second card is 6. There is no second option or second case here, so it counts only once.

Samantha22 · Answer

I like the solution - it’s helpful.

Nenye_17 · Answer

I like the solution - it’s helpful.

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