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M05-10

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New post 16 Sep 2014, 00:24
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Question Stats:

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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. Then another card is drawn and put back. If the sum of the two cards 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. \(\frac{1}{3}\)

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New post 16 Sep 2014, 00:24
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Official Solution:

There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards 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. \(\frac{1}{3}\)


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: \(P = \frac{2}{5}\).


Answer: D
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New post 01 Nov 2016, 05:15
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uvemdesalinas wrote:
Bunuel wrote:
Official Solution:

There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards 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. \(\frac{1}{3}\)


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: \(P = \frac{2}{5}\).


Answer: D


Bunuel,

If cards are numered from 1 to 6, how can the scenario (4,4) be possible?
Once you have retried the 4 card you would not have the option to retry another 4...

Thanks
V


Please read carefully: There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back.
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New post 02 Jan 2015, 20:45
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.
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New post 05 Jan 2015, 05:44
munukutlask wrote:
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.


(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.
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New post 20 Feb 2015, 05:16
Bunuel

If we are not considering 1,7 as a case than that means that Ace can never be considered as 1. But is that a rule for the card questions in GMAT?
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New post 20 Feb 2015, 06:09
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New post 31 Mar 2015, 05:40
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?
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New post 31 Mar 2015, 06:02
DarkMight wrote:
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?


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.
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New post 07 Nov 2015, 00:42
Dear Bunuel,

Just wondering.. if we had to choose 2 cards from 6 cards and did not replace the card. What is the probability of one of the cards is 5? Thank you!
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New post 14 Oct 2016, 10:50
Liza99:
Case 1 := 1st card is 5 and second card is any other card out of 1,2,3,4,6 Pr = (1/6)
OR
Case 2 := First card is any card but 5 and second card is 5 , Pr = (5/6)*(1/5) = 1/6

So total probability = case 1 + case 2 = 1/3
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New post 01 Nov 2016, 05:13
Bunuel wrote:
Official Solution:

There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards 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. \(\frac{1}{3}\)


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: \(P = \frac{2}{5}\).


Answer: D


Bunuel,

If cards are numered from 1 to 6, how can the scenario (4,4) be possible?
Once you have retried the 4 card you would not have the option to retry another 4...

Thanks
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New post 22 Sep 2018, 09:06
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.
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New post 23 Sep 2018, 04:02
stark026arya wrote:
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.


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.
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New post 11 Nov 2018, 15:16
Bunuel wrote:
stark026arya wrote:
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.


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.


Hi Bunuel,

Thanks for the explanation. Can you please provide links to a few official questions that test the conditional probability. Also, if there is any article/resource to learn more about this concept.

I did the exact same thing that poster above did. Just calculated the 1/18 probability.

Thanks much!
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New post 12 Nov 2018, 00:08
thebest wrote:
Bunuel wrote:
stark026arya wrote:
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.


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.


Hi Bunuel,

Thanks for the explanation. Can you please provide links to a few official questions that test the conditional probability. Also, if there is any article/resource to learn more about this concept.

I did the exact same thing that poster above did. Just calculated the 1/18 probability.

Thanks much!


Search in OG probability questions.

Official Guide DS probability questions: https://gmatclub.com/forum/search.php?s ... mit=Search

Official Guide PS probability questions: https://gmatclub.com/forum/search.php?s ... mit=Search

For more on the subject check the links below:

22. Probability



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New post 12 Nov 2018, 03:14
Bunuel wrote:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards 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. \(\frac{1}{3}\)



combination whose sum =8 are (2,6),(3,5),(4,4),(5,3),(6,2)

so P of getting 5 is = 2/5 option D
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New post 16 Nov 2018, 17:55
Hey Bunuel

I arrived at correct answer, but slightly different from you approach.

I listed the pairs like you have i.e (6,2) (2,6) (5,3) (3,5) (4,4) - only 5 possible scenarios for sum to be 8.

However I chose individual numbers, non repeating ones i.e 6,2,5,3,4 and not the pairs.

so 5 numbers and two picks w/o replacement: so 2/5 is this correct ?
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New post 01 Mar 2019, 01:35
I think this is a high-quality question and I agree with explanation.
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New post 15 Jun 2019, 21:18
Bunuel wrote:
There are 6 cards numbered from 1 to 6. They are placed into a box, and then one is drawn and put back. Then another card is drawn and put back. If the sum of the two cards 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. \(\frac{1}{3}\)


Hello Bunuel,
Could you please point out mistake in my interpretation:
(Probability of drawing a 5 and a 3 ) * 2 ( for reverse of it)
1/6 * 1/6 * 2 = 1/18
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Re: M05-10   [#permalink] 15 Jun 2019, 21:18

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