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# M05-10

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Director
Joined: 22 Nov 2018
Posts: 502
Location: India
GMAT 1: 640 Q45 V35
GMAT 2: 740 Q49 V41

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15 Jun 2019, 22:23
pm0103 wrote:
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

pm0103 You are ignoring the fact that the stem says sum of the numbers picked is 8. You are trying to arrive at the prob of picking 3 and 5 initially from 1-6 cards with repetition, where as the question says already the cards are picked and you have a sum of 8(4,4/6,2/2,6/3,5/5,3) and requires what is the prob of one of the cards being a 5. Which is 2/5.

Hope this helps!

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Joined: 04 Jun 2019
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13 Sep 2019, 19:19
if cards are numbered as 1 to 6 , then according to my understanding 4,4 is not possible , however case has been considered in answer explanations. please advice. Thanks
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Joined: 02 Sep 2009
Posts: 64172

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14 Sep 2019, 04:46
princesingla1986 wrote:
if cards are numbered as 1 to 6 , then according to my understanding 4,4 is not possible , however case has been considered in answer explanations. please advice. Thanks

Check the highighted part:

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?

The card is drawn and the put back, so it's possible to draw 4, put it back, and then draw 4 again.
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24 Jan 2020, 02:39
Bunuel,

Why do we care about two different cases of 5,3 and 6,2 if in sum they give 8? any of the pair is eligible to bear 5 or 3 and 6 or 2.
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Joined: 02 Sep 2009
Posts: 64172

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24 Jan 2020, 05:22
nurba92 wrote:
Bunuel,

Why do we care about two different cases of 5,3 and 6,2 if in sum they give 8? any of the pair is eligible to bear 5 or 3 and 6 or 2.

(5, 3) and (3, 5) are 2 different cases:
First draw = 5 and second draw = 3;
First draw = 3 and second draw = 5.
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Concentration: General Management, Technology
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18 May 2020, 13:40
Alternative solution:

Since sum is 8, we must distribute 2 d either side of spacer |
think about permutation of 4 d and one spacer |
dddd|
total number of permutations = 5!/4! = 5
Also sum is 8, so if one card is 5 the other must be 3. And this is total 2 occurrences, as 5 can appear in the first draw or the second draw.
Required probability = 2/5

Re: M05-10   [#permalink] 18 May 2020, 13:40

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