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# 6 cards are numbered with 1, 2,3, 4, 5, and 6. Firstly, one

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6 cards are numbered with 1, 2,3, 4, 5, and 6. Firstly, one [#permalink]

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02 Aug 2006, 20:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

6 cards are numbered with 1, 2,3, 4, 5, and 6. Firstly, one cards was picked up and replaced back, then, the second cark was picked up randomly. If the sum of the numbers in the 2 cards is 8, what is the probability that one of the cards has number 5?
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02 Aug 2006, 20:49
2/5
Total possibilities
4,4
2,6
6,2
3,5
5,3
5 is there in 2 out of total 5.
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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02 Aug 2006, 20:58
Number of ways two cards are picked such that ..sum is 8

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

That is there are 5 possibilties.

There are two ways that 5 can be in the sum 3,5 and 5,3

hence probability = 2/5
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03 Aug 2006, 01:05
I think it is 2 : 4. Maybe I missed something, but how can 4 and 4 be a possiblitly when there is only one 4?
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03 Aug 2006, 01:06
Ignore the post above.
Senior Manager
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03 Aug 2006, 04:33
Thanks again guys. You came thru for me again
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03 Aug 2006, 23:57
Ways you can get 8:
2,6
6,2
3,5
5,3
4,4

Ways including 5: 2
Prob = 2/5
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question bank for probability questions [#permalink]

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04 Aug 2006, 11:37
can someone please point me to the question bank of probablity questions ?
Appreciate it !
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Winsome

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Re: question bank probability question [#permalink]

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05 Aug 2006, 20:01
apollo168 wrote:
6 cards are numbered with 1, 2,3, 4, 5, and 6. Firstly, one cards was picked up and replaced back, then, the second cark was picked up randomly. If the sum of the numbers in the 2 cards is 8, what is the probability that one of the cards has number 5?

Prob of picking 2 cards = 6c2 = 15
Prob of favourable events = {(3,5)}

Not sure if order matters

Prob = 1/15

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06 Aug 2006, 11:53
ps_dahiya wrote:
2/5
Total possibilities
4,4
2,6
6,2
3,5
5,3
5 is there in 2 out of total 5.

hi Ps,

Can the question be ineterpreted as follows

5, 3 and 3, 5 are the two options if the sum is the 8

prob = 2/(6x6) =1/18

why do we conclude that the prob is among all combos of 8 as sum ?

thanks
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06 Aug 2006, 19:00
old_dream_1976 wrote:
ps_dahiya wrote:
2/5
Total possibilities
4,4
2,6
6,2
3,5
5,3
5 is there in 2 out of total 5.

hi Ps,

Can the question be ineterpreted as follows

5, 3 and 3, 5 are the two options if the sum is the 8

prob = 2/(6x6) =1/18

why do we conclude that the prob is among all combos of 8 as sum ?

thanks

The bold part will be right is asked:
what is the prob that sum is 8 AND one of the cards bears a 5.

prob of having one card as 5 when the sum of two cards already drawn is 8.
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06 Aug 2006, 19:41
ps_dahiya wrote:
old_dream_1976 wrote:
ps_dahiya wrote:
2/5
Total possibilities
4,4
2,6
6,2
3,5
5,3
5 is there in 2 out of total 5.

hi Ps,

Can the question be ineterpreted as follows

5, 3 and 3, 5 are the two options if the sum is the 8

prob = 2/(6x6) =1/18

why do we conclude that the prob is among all combos of 8 as sum ?

thanks

The bold part will be right is asked:
what is the prob that sum is 8 AND one of the cards bears a 5.

prob of having one card as 5 when the sum of two cards already drawn is 8.

6 cards are numbered with 1, 2,3, 4, 5, and 6. Firstly, one cards was picked up and replaced back, then, the second cark was picked up randomly. up untill here the question says cards were drawn one after the other . If the sum of the numbers in the 2 cards is 8, (does not say in all such card drawing instances.). what is the probability that one of the cards has number 5?

I see your point but I dont think this problem is worded properly. I hope we do not see such vague questions.

But thanks for the response.
06 Aug 2006, 19:41
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