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Probability

by MBA.Aspirant » Wed Jun 22, 2011 10:10 am
From the even numbers between 1 and 9, two different even numbers are to be choosen at random. what is the probability that their sum will be 8?

a)1/6 b)3/16 c)1/4 d)1/3 e)1/2

we have 2 events: 2+6, 6+2

my question is how to know quickly the number of summing events which =12?
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by Frankenstein » Wed Jun 22, 2011 10:16 am
Hi,
There are 4 even numbers between 1 and 9
We can pick 1st number from 4 in 4 ways
2nd number from the remaining 3 in 3 ways
So, total 4.3 = 12 ways
Essentially we are picking two from 4 and order matters. So, 4P2 = 4!/2! = 12
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by MBA.Aspirant » Wed Jun 22, 2011 10:18 am
Thanks! I was just about to write the massive discovery that is 3*4 :)
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by manpsingh87 » Wed Jun 22, 2011 10:23 am
MBA.Aspirant wrote:From the even numbers between 1 and 9, two different even numbers are to be choosen at random. what is the probability that their sum will be 8?

a)1/6 b)3/16 c)1/4 d)1/3 e)1/2

we have 2 events: 2+6, 6+2

my question is how to know quickly the number of summing events which =12?
there are 4 even nos. between 1 and 8, these are 2,4,6,8 out of these even nos any two can be selected in 4c2 ways=6,
now, sum will be eight if the selected numbers are 2 and 6,

so for first draw we can have two options either it will be 2 or 6 and for the second draw we have only one option left i.e. if 2 is drawn in the first draw then the second must have 6 and if 6 is drawn in the first draw then the second draw must have 2.

hence required probability is 2/6=1/3 hence D
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by Frankenstein » Wed Jun 22, 2011 10:27 am
manpsingh87 wrote:
there are 4 even nos. between 1 and 8, these are 2,4,6,8 out of these even nos any two can be selected in 4c2 ways=6,
now, sum will be eight if the selected numbers are 2 and 6,

so for first draw we can have two options either it will be 2 or 6 and for the second draw we have only one option left i.e. if 2 is drawn in the first draw then the second must have 6 and if 6 is drawn in the first draw then the second draw must have 2.

hence required probability is 2/6=1/3 hence D
Hi,
In order to calculate total cases, you have considered selection(without order) but, for the favorable case you considered order. It doesn't work that way.
2 numbers from 4 evens is picked in 4C2 = 6
required pair(2,6) can be selected in 1 way.
So, probability is 1/6
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by manpsingh87 » Wed Jun 22, 2011 7:56 pm
Frankenstein wrote:
manpsingh87 wrote:
there are 4 even nos. between 1 and 8, these are 2,4,6,8 out of these even nos any two can be selected in 4c2 ways=6,
now, sum will be eight if the selected numbers are 2 and 6,

so for first draw we can have two options either it will be 2 or 6 and for the second draw we have only one option left i.e. if 2 is drawn in the first draw then the second must have 6 and if 6 is drawn in the first draw then the second draw must have 2.

hence required probability is 2/6=1/3 hence D
Hi,
In order to calculate total cases, you have considered selection(without order) but, for the favorable case you considered order. It doesn't work that way.
2 numbers from 4 evens is picked in 4C2 = 6
required pair(2,6) can be selected in 1 way.
So, probability is 1/6
yep..i forgot to multiply 4C2 with 2!,,,thats what happen when you try to reply in an impromptu manner..!!!!
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by winniethepooh » Wed Jun 22, 2011 9:41 pm
Hey MBA aspirant what was your question?
I mean, what did you really want to know? I didn't get it.
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