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Probality

by alltimeacheiver » Sat Apr 30, 2011 4:10 am
ques

A=(2,3,4,5)
B=(4,5,6,7,8)

two integers will be randomly selected from the sets above one integers set A and One integer set from b. What is the probality that summs of inteers will be equal to 9

I can make combination 4 like, (4,5), (5,4), (3,6), (2,7)

So the probality will 4/20= 0.2


I want to ask gmat guns that in multiplication cases we take reverse of these possibilities. suppose in same problem set lets take that we found 4 cases whuch make even possibility. (4,5) and (5,4), )2,7), (7,2), (3,6),(6,3) and so on This just an example. So in Mulptiplication we take 8 cases that is double of it.
In addition we count one and in mulptiplication we count twice depending upon no of sets.

Pls confirm
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by vineeshp » Sat Apr 30, 2011 4:33 am
I dont think I understand your question.

You cannot take cases which dont exist.
For eg, if there is no 7 in the first set, how can you take 7,2? You cant.

But I really dunno if I followed your question.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by pemdas » Sat Apr 30, 2011 7:39 am
@alltimeacheiver - your probability stands to be the same 1/5

if you do permutation (that's what you meant perhaps) and count set1-set2 then set2-set1 your favorable outcomes will be 8 (you correctly found this) BUT your total possible outcomes should be 40 not 20 then. So 8 out of 40 makes 1/5 again. This will help you prove that the order does not matter here, 9 is 9 and the changes in selection sequence (from set1 to set2 or vice verse) will not affect the probability. The confusion with your multiplication decision tree occurs when you treat the favorable outcomes as permutations and the total possible outcomes as combination. We should treat both one way, or we will miscount the probability.
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by venmic » Sun May 01, 2011 2:11 pm
what is the answer to this one

4/20 since the combinations are (2,7) (3,6)(4,5)(5,4)

or

3/20 since (4,5) and (5,4) is the same combination
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by srcc25anu » Sun May 01, 2011 2:50 pm
IMO it should be 1/5

cases when sum of integers = 9 are (2,7) (3,6) (4,5) and (5,4) so 4 cases
total ways to select 2 integers out of 9 taking 1 from set A and 1 from set B = 4c1 * 5c1 = 4*5 = 20

hence prob = 4/20 = 1/5
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by jcc0523 » Sun May 01, 2011 6:14 pm
srcc25anu wrote:IMO it should be 1/5

cases when sum of integers = 9 are (2,7) (3,6) (4,5) and (5,4) so 4 cases
total ways to select 2 integers out of 9 taking 1 from set A and 1 from set B = 4c1 * 5c1 = 4*5 = 20

hence prob = 4/20 = 1/5

Does multiplying the total number in set 1 by the total number in set 2 equal the total number of ways to to select two integers which equal 9?
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by srcc25anu » Sun May 01, 2011 8:42 pm
to jcc0523:

yes, total ways to select 1 from set A and 1 from set B = 4C1 * 5C1 = 4*5 = 20
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