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Intern  Joined: 01 Nov 2015
Posts: 4
Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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2
Hi,

Sorry for bring up this post. I am still not clear about the answer. May I explain my understanding:
There are 6 outcomes for the first draw, 6 for the seconds. So total is 36.
Still there are 5 outcomes that sum is 8: (2;6)(6;2)(3;5)(5;3)(4;4) ; in which, two of them (3;5)(5;3) have "5"
So the possibility to have "5" is (5/36)*(2/5)=1/18

And my question is:
Since Bunuel said 5 outcomes [which is (2;6)(6;2)(3;5)(5;3)(4;4)] is all possibilities we have, why not 36?
Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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1
McQueen wrote:
Hi,

Sorry for bring up this post. I am still not clear about the answer. May I explain my understanding:
There are 6 outcomes for the first draw, 6 for the seconds. So total is 36.
Still there are 5 outcomes that sum is 8: (2;6)(6;2)(3;5)(5;3)(4;4) ; in which, two of them (3;5)(5;3) have "5"
So the possibility to have "5" is (5/36)*(2/5)=1/18

And my question is:
Since Bunuel said 5 outcomes [which is (2;6)(6;2)(3;5)(5;3)(4;4)] is all possibilities we have, why not 36?

Hi,
the Q could be asked in two ways --
1) Prob that the Sum of two cards is 8 and one card is 5..
totao ways here will be 6*6 and answer will be 2/36 =1/18
2) prob that one card is 5 when the SUM is 8..
this is what is being asked here. we are already sure that the SUM is 8 and we have to work further on this info
so total ways of 8 are 2,6 : 6,2: 3,5 : 5,3 : 4,4
so we are working on thi sand our prob is to find how many of these ways have 5 as a card..

Hope it clears the Query

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Intern  S
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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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Here's how i solved this problem.
We are drawing 2 cards and the sum is 8 according to the question. So we just have to find out how many cases we have when the sum on the cards is 8. We should not be bothered about the total outcomes when drawing 2 cards out of 6 cards.

So we have : 2+6, 3+5, 4+4, 5+3, 6+2

Out of these 5 scenarios we need to find that how many of these 5 cases have 5 numbered card in one of the cards, which is 2 scenarios i.e 3+5 and 5+3

Thus 2 out of 5 cases satisfy the conditions given in question. Thus answer is 2/5.
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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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Hi Bunuel
Do not we need to get the total number of card selected randomly and then get the probability of #5 card.
I mean 6 card first time and another 6 cards second time. thus we have 36 possible cards to be selected.
the final step is to divide 2 possible of #5 cards so 2 divided 36 = 1/18.
I know OA is 2/5 which is right. but I am telling what is running in my mind.
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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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hatemnag wrote:
Hi Bunuel
Do not we need to get the total number of card selected randomly and then get the probability of #5 card.
I mean 6 card first time and another 6 cards second time. thus we have 36 possible cards to be selected.
the final step is to divide 2 possible of #5 cards so 2 divided 36 = 1/18.
I know OA is 2/5 which is right. but I am telling what is running in my mind.

No, you are missing the case when the sum of the 2 cards chosen = 8. This will reduce your number of cases allowed to 5 from 36. Had the question asked you numbers of cases possible with 1 number as 5 (with no restriction on the sum), then your answer should have been 2/36 or 1/18.

The added restriction of the the sum of the 2 cards drawn = 8 reduces the allowed cases to 5.

Hope this helps.
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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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Total ways of drawing 2 cards with replacement = 6*6 = 36
We need the sum to be 8. It is possible in the following ways:

(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)
Hence a total of 5 ways.
Out of these, only in two ways do we get a 5.

Hence the required probability = 2/5

Correct Option: D
Manager  Joined: 12 Jan 2015
Posts: 196
Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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Quote:

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=2/5.

Hope it's clear.

Hi Bunuel / chetan2u,

I agree with this approach but this didn't strike in my mind while attempting question .
I followed this approach but didn't got the required answer. Can you please suggest what is wrong.

We have 6 cards numbering 1,2,3,4,5,6 out of this we need to select 2 card sum of which is 8 and find the probability for the occurrence of No5.
So, the only possibility we have is (3,5) or (5,3)

Selecting 3 out of 6 numbers= 1/6
Selecting 5 out of 6 numbers= 1/6

1/6 * 1/6= 1/36

Multiply by 2 as there are two conditions (3,5) and (5,3)= (1/36)*2= 1/18

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Prakhar
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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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1
PrakharGMAT wrote:
Quote:

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=2/5.

Hope it's clear.

Hi Bunuel / chetan2u,

I agree with this approach but this didn't strike in my mind while attempting question .
I followed this approach but didn't got the required answer. Can you please suggest what is wrong.

We have 6 cards numbering 1,2,3,4,5,6 out of this we need to select 2 card sum of which is 8 and find the probability for the occurrence of No5.
So, the only possibility we have is (3,5) or (5,3)

Selecting 3 out of 6 numbers= 1/6
Selecting 5 out of 6 numbers= 1/6

1/6 * 1/6= 1/36

Multiply by 2 as there are two conditions (3,5) and (5,3)= (1/36)*2= 1/18

Hi,
what is the probability of picking two numbers such that their sum is 8 and one of the card is 5?..

But the Q tells you that the two cards have been already taken out such that their sum is 8..
Now we have to check the probability that if the sum is 8, what is the probability that one of the cards is 5..
so we work ahead of the point that sum is 8..

ways it can be 8 is 2,6 or 6,2 ; 3,5 or 5,3 ; and 4,4...
so toal ways the sum can be 8 is 5 ways and TWO of them have 5 as a card..
so Probability = 2/5
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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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Hi chetan2u,

Thank you so much. Now I got the point where I was lacking.
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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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Restated Question: Of the pairs of numbers that sum to 8, what is the probability a pair of numbers will contain a '5'.

-- Only way to guarantee a sum of 8 is to first draw either a 2,3,4,5 or 6. This is your set of total possibilities
-- Each number from the set {2,3,4,5,6} is paired with another number that together sum to 8, so we now look for how many of these pairs contain the number 5
-- We find that drawing a '3' and a '5' on the first draw each guarantee us a card numbered '5'

In summary:
2 pairs contain the number '5'
5 total pairs of numbers that sum to '8'

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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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The cards add up to 8. There are five possibilities (first number is first card drawn; second number is second card drawn):

2, 6
6, 2
3, 5
5, 3
4, 4

2 out of the 5 possibilities have a 5. 2/5.

Kudos if you agree!! Got a better method?
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Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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ggarr wrote:
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5?

A. 2/5
B. 3/5
C. 1/18
D. 1/3
E. 1/6

1. The total number of cases is restricted by the fact that the sum of the numbers of the cards drawn be 8. So total number of cases instead of being 6*6 will be only those where the sum of the numbers is 8. The cases are (2,6), (6,2), (5,3), (3,5), (4,4)
2. The favorable outcome is in two such cases (5,3) and (3,5)
3.So the probability is 2/5
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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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This problem becomes much easier if you insert yourself into the situation. Pretend you have physically dropped 6 cards into an empty bowl, and you select one card, look at it, drop that card back into the bowl and select another card. If the sum of your card selections is 8, how many different scenarios can this occur?

Pick the 2, drop it in the bowl, select the 6.
Pick the 6, drop it in the bowl, select the 2.
Pick the 3, drop it in the bowl, select the 5.
Pick the 5, drop it in the bowl, select the 3.
Pick the 4, drop it in the bowl, select the 4 again.

In the real world, how could you have two different scenarios where you pick the 4 first, drop it back in the bowl, and pick that same 4 again? You can't!

Therefore, there are 5 different choices, and 2 of those choices include the card with a 5.

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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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I was long confused why is 2/5 and not 2/10 since there are 10 potential cards and only two 5's. So for who might be confused on the same thing as me you can also approach it by subtracting from 1 the probability of no 5th. and we have 3 scenarios with no 5's so 6 cards out of 10. 1-6/10=4/10 or 2/5
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Re: Six cards numbered from 1 to 6 are placed in an empty bowl.  [#permalink]

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_________________ Re: Six cards numbered from 1 to 6 are placed in an empty bowl.   [#permalink] 22 Jul 2018, 13:38

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