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# M05-10

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Math Expert
Joined: 02 Sep 2009
Posts: 64068

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15 Sep 2014, 23:24
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55% (hard)

Question Stats:

55% (01:34) correct 45% (01:49) wrong based on 399 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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15 Sep 2014, 23:24
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1
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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Joined: 16 May 2011
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02 Jan 2015, 19:45
2
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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20 Feb 2015, 04:16
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?
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Joined: 02 Sep 2009
Posts: 64068

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01 Nov 2016, 04:15
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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23 Sep 2018, 03:02
1
1
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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05 Jan 2015, 04:44
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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20 Feb 2015, 05:09
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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Joined: 14 Feb 2013
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31 Mar 2015, 04:40
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?
Math Expert
Joined: 02 Sep 2009
Posts: 64068

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31 Mar 2015, 05:02
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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06 Nov 2015, 23:42
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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14 Oct 2016, 09:50
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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Joined: 10 Oct 2016
Posts: 8

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01 Nov 2016, 04:13
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: $$P = \frac{2}{5}$$.

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
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Joined: 18 Feb 2018
Posts: 3

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22 Sep 2018, 08:06
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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Posts: 7
Location: United States
Schools: Kellogg '21

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11 Nov 2018, 14:16
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,

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

Thanks much!
Math Expert
Joined: 02 Sep 2009
Posts: 64068

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11 Nov 2018, 23:08
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,

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

Thanks much!

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12 Nov 2018, 02:14
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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16 Nov 2018, 16:55
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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01 Mar 2019, 00:35
I think this is a high-quality question and I agree with explanation.
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Joined: 12 Nov 2016
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15 Jun 2019, 20:18
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, 20:18

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# M05-10

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