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Math Expert V
Joined: 02 Sep 2009
Posts: 59588

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2
10 00:00

Difficulty:   55% (hard)

Question Stats: 55% (01:33) correct 45% (01:51) wrong based on 217 sessions

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 59588

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3
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: $$P = \frac{2}{5}$$.

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59588

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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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Location: United States
Concentration: Entrepreneurship, Marketing
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1
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?
Math Expert V
Joined: 02 Sep 2009
Posts: 59588

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qw1981 wrote:
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?

Those are just cards numbered from 1 to 6, not playing cards. There are no aces there.
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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 59588

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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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Intern  B
Joined: 26 Oct 2014
Posts: 22

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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!
Intern  Joined: 23 Apr 2016
Posts: 21
Location: Finland
GPA: 3.65

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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
Intern  Joined: 10 Oct 2016
Posts: 8

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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: $$P = \frac{2}{5}$$.

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
Math Expert V
Joined: 02 Sep 2009
Posts: 59588

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1
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: $$P = \frac{2}{5}$$.

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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Joined: 18 Feb 2018
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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59588

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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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Joined: 20 Apr 2018
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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,

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

Thanks much!
Math Expert V
Joined: 02 Sep 2009
Posts: 59588

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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,

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

Thanks much!

Search in OG probability questions.

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For more on the subject check the links below:

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GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5436
Location: India
Concentration: Sustainability, Marketing
GPA: 4
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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
Manager  G
Status: Studying 4Gmat
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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 ?
Director  G
Joined: 22 Nov 2018
Posts: 562
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GMAT 1: 640 Q45 V35 GMAT 2: 660 Q48 V33 ### Show Tags

I think this is a high-quality question and I agree with explanation.
Intern  B
Joined: 12 Nov 2016
Posts: 7

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1
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 Re: M05-10   [#permalink] 15 Jun 2019, 21:18

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# M05-10

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