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M05-10

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M05-10  [#permalink]

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New post 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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New post 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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New post 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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Re: M05-10  [#permalink]

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New post 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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New post 24 Jan 2020, 05:22
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Re: M05-10  [#permalink]

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New post 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

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

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