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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 one of the cards was 5 if the sum of the two was 8?

The probability is \(\frac{2}{5}\) . You have the possibilities of 2,6 3,5 4,4 5,3 6,2 Two out of these five combinations have 5's in them, so the probability that you have a 5 is 2/5.

I counted the event {4,4} twice in the calculation, which makes a probability of 2/6=1/3 so E. Even if they are the same cards, I don't understand why we can't make a permutation of the cards #4, i.e. counting it twice. Can you please tell me where I am wrong?

I did it this way - Probability of the picking the first card - 5/6 -- ( The cards must either by 2,4,5,6,3)

Probability of picking second card - 1/6 - Because if you select any of the five cards in the first card , you can pick only its complement in the second card - Eg: 2 must have 6

So - 5/6 * 1/6 = 5/36

I am still confused where I made a mistake - Could some one help me out ?

What do you mean when you say that "the probability of picking the first card is 5/6"? 5/6 is the probability of picking any of the five particular cards from 6. For example 5/6 is the probability of picking either 2, 3, 4, 5, or 6. But why do you consider this case? The same question for the second card.

Anyway, we are told that two cards were selected and the sum of these two cards is 8 (this scenario has already happened). The question then asks about the probability that one of the cards was 5. There are five cases possible the sum to be 8: (6,2); (2,6); (5,3); (3,5); (4,4).

One from this 5 has already happened. From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.

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 one of the cards was 5 if the sum of the two was 8? (A) 1 (B) \(\frac{2}{3}\) (C) \(\frac{1}{2}\) (D) \(\frac{2}{5}\) (E) \(\frac{1}{3}\)

Can somebody will try to explain this problem with union and intersection method. Not through permutation and combination. probability of getting 5 in first draw is 1/6, then in second draw outcome must be 3. How we will calculate this must be event. The probability of getting 3 in the second draw is also 1/6, but I think we need to calculate here must be event dependent on the first event.

Isn't the phrasing of this question misleading? Possibly it's just me because it seems that everyone else got it, but would someone pls explain to me how I'm reading this question wrong?

Since it asks "What is the probability that one of the cards was 5 if the sum of the two was 8?", the condition "if the sum of the two was 8" implies only 2 situations that this can occur. Then if this condition is true, "What is the probability that one of the cards was 5?". In both situations (i.e. 3+5, 5+3), it can only happy 2/4 times or 1/2, so shouldn't the answer be C?

If the question asked "what is the probability of drawing a combination which sums to 8 if one card is 5?", then I would agree, 2/5 is correct.

Isn't the phrasing of this question misleading? Possibly it's just me because it seems that everyone else got it, but would someone pls explain to me how I'm reading this question wrong?

Since it asks "What is the probability that one of the cards was 5 if the sum of the two was 8?", the condition "if the sum of the two was 8" implies only 2 situations that this can occur. Then if this condition is true, "What is the probability that one of the cards was 5?". In both situations (i.e. 3+5, 5+3), it can only happy 2/4 times or 1/2, so shouldn't the answer be C?

If the question asked "what is the probability of drawing a combination which sums to 8 if one card is 5?", then I would agree, 2/5 is correct.

Where am I going wrong??

"What is the probability that one of the cards was 5 if the sum of the two was 8?", I think, sum of 8 can be achieved by all the five possible combination. First, you have to fulfill the if clause after that only you can think about then clause. i.e if sum was 8 then what is the prob of 5.

Isn't the phrasing of this question misleading? Possibly it's just me because it seems that everyone else got it, but would someone pls explain to me how I'm reading this question wrong?

Since it asks "What is the probability that one of the cards was 5 if the sum of the two was 8?", the condition "if the sum of the two was 8" implies only 2 situations that this can occur. Then if this condition is true, "What is the probability that one of the cards was 5?". In both situations (i.e. 3+5, 5+3), it can only happy 2/4 times or 1/2, so shouldn't the answer be C?

If the question asked "what is the probability of drawing a combination which sums to 8 if one card is 5?", then I would agree, 2/5 is correct.

Where am I going wrong??

"What is the probability that one of the cards was 5 if the sum of the two was 8?", I think, sum of 8 can be achieved by all the five possible combination. First, you have to fulfill the if clause after that only you can think about then clause. i.e if sum was 8 then what is the prob of 5.

------- Oh ok thanks - seems so obvious now once you put it that way. I was reading it the other way around which was leading me to think of 2 sets instead of 5 sets. Thanks for the clarification.

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 one of the cards was 5 if the sum of the two was 8?

The probability is \(\frac{2}{5}\) . You have the possibilities of 2,6 3,5 4,4 5,3 6,2 Two out of these five combinations have 5's in them, so the probability that you have a 5 is 2/5.

I counted the event {4,4} twice in the calculation, which makes a probability of 2/6=1/3 so E. Even if they are the same cards, I don't understand why we can't make a permutation of the cards #4, i.e. counting it twice. Can you please tell me where I am wrong?

Thanks in advance .

Probablity is like a dreaded subject...even the simplest problem(now) looks like a major hurdle.....

thanks for the nice explanation.... _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Why are we considering (5,3) and (6,2) again after having considered (3,5) and (2,6)? When the card after first draw is put back, how does it matter whether 5 appears in the first draw or the second?

If possible, please explain with a simple example. Thanks.

Why are we considering (5,3) and (6,2) again after having considered (3,5) and (2,6)? When the card after first draw is put back, how does it matter whether 5 appears in the first draw or the second?

If possible, please explain with a simple example. Thanks.

here actually the arrangement is important. You can see it this way

you first draw a 5 u put it back and then another pick gives 3 you now have 8 now in second go you first get 3 and then 5 =>8

the thing here is, in how many events/ actions you can have 8, whether you pick 5 first or 3 both matters because it counts for the number of events/ actions performed to get 8.

hope this clarifies..!! _________________

Practice Practice and practice...!!

If my reply /analysis is helpful-->please press KUDOS If there's a loophole in my analysis--> suggest measures to make it airtight.

Why are we considering (5,3) and (6,2) again after having considered (3,5) and (2,6)? When the card after first draw is put back, how does it matter whether 5 appears in the first draw or the second?

If possible, please explain with a simple example. Thanks.

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 one of the cards was 5 if the sum of the two was 8? (A) 1 (B) \(\frac{2}{3}\) (C) \(\frac{1}{2}\) (D) \(\frac{2}{5}\) (E) \(\frac{1}{3}\)

What combinations of two cards are possible to total 8? (first card, second card): (6,2) (2,6) (5,3) (3,5) (4,4) – only 5 possible scenarios for sum to be 8. One from this 5 has already happened. Notice here that the case of (6,2) is different from (2,6), and similarly the case of (5,3) is different from (3,5).

From this five cases, only in two we have 5. So, the probability is 2 chances out of 5 that the one that occurred had 5: P=2/5.

I did it this way - Probability of the picking the first card - 5/6 -- ( The cards must either by 2,4,5,6,3)

Probability of picking second card - 1/6 - Because if you select any of the five cards in the first card , you can pick only its complement in the second card - Eg: 2 must have 6

So - 5/6 * 1/6 = 5/36

I am still confused where I made a mistake - Could some one help me out ?

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 one of the cards was 5 if the sum of the two was 8?

The probability is \(\frac{2}{5}\) . You have the possibilities of 2,6 3,5 4,4 5,3 6,2 Two out of these five combinations have 5's in them, so the probability that you have a 5 is 2/5.

I counted the event {4,4} twice in the calculation, which makes a probability of 2/6=1/3 so E. Even if they are the same cards, I don't understand why we can't make a permutation of the cards #4, i.e. counting it twice. Can you please tell me where I am wrong?

Thanks in advance .

I went the long way to solve this problem.

The question is looking for a combination of 3 and 5.

Step 1: In the first draw what are the chances of drawing a 3 or a 5? Simple. There are 6 options and the chances that any of the two will come up is 2/6 = 1/3.

Step 2: Once again we have 6 options but one of 3 or 5 must have been chosen on the way to getting an 8. So we need to draw 5 if 3 had been chosen or draw 3 if 5 had been chosen. This time around we go the combinatorics way. 3 or 5 have become one option. Therefore the chances of choosing either of them becomes 6!/2!4! = 1*2*3*4*5*6/1*2*1*2*3*4 = 15. Flip this over and you get a 1/15 chance of choosing 3 or 5, if one had been chosen before.

Step 3: This is an AND decision in which both Steps 1 and 2 must occur so we can have a 5 and a 3. That means we add 1/3 to 1/15. 1/3 + 1/15 = 2/5. _________________

I don't quite get why people count 5,3 and 3,5 as different combinations. The question says nothing about order. The way I reason this, you can have the following combinations

cards 6 and 2

5 and 3

4 and 4

So if one of the cards is 5, the other is 3, regardless of the order that they were pulled out of the stack. that's 1 out 3 cards, no?

I don't quite get why people count 5,3 and 3,5 as different combinations. The question says nothing about order. The way I reason this, you can have the following combinations

cards 6 and 2

5 and 3

4 and 4

So if one of the cards is 5, the other is 3, regardless of the order that they were pulled out of the stack. that's 1 out 3 cards, no?

We have two cards, thus the case (first card)=5 and (second card)=3 is different from (first card)=3 and (second card)=5.