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Combinatorics Question Veritas

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Combinatorics Question Veritas

by Perminology » Wed Nov 25, 2009 10:26 pm
A=(2,3,4,5)
B=(4,5,6,7,8)

Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers equals 9?

A) 0.15
B) 0.20
C) 0.25
D) 0.30
E) 0.33

Will show OA after a reply. I want to see what you guys think first.
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by papgust » Wed Nov 25, 2009 11:29 pm
IMO it is B.

We have total combinations of 20 - (A * B = 5 * 4)

To satisfy the restriction (sum=9) - (2,7), (3,6), (4,5), (5,4) = 4 combinations

Probablity of sum of a combination that equals 9 = 4/20 = 0.20
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by Perminology » Thu Nov 26, 2009 12:11 am
Interesting... wouldn't you count the other two as well?
(6,3) (7,2)? Does order matter here?
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by papgust » Thu Nov 26, 2009 1:02 am
I missed (6,3) and (7,2). My mistake.

Yes, you should include these 2 as well. Then it will 6/20 = 0.30 (D)
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by viju9162 » Thu Nov 26, 2009 1:28 am
What is the OA ?
"Native of" is used for a individual while "Native to" is used for a large group
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by thephoenix » Thu Nov 26, 2009 2:35 am
IMO A
total no of cases=40
desirable cases=6
p=6/40=.15
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by Perminology » Thu Nov 26, 2009 5:34 am
And the OA is B!
papgust you were right the first time. The book doesn't count the last two, so I posted here to see what others thought. I'm still confused as to whether or not it's double counting to add the last two combinations or if there is a problem with the book. Could anyone else help to clarify?

Thanks in advance!

[Edit: I have caught some Veritas' workbooks to have some typos in answer explanations so I posted the question here]
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by papgust » Thu Nov 26, 2009 6:43 am
Initially i felt that it should be B. But even now i feel that it should be D instead.

Take an example here:
If Two dices are thrown simultaneously, what is the probability that the sum of the numbers shown in both dices is 3?

What are the possible cases here?

(2,1), (1,2) --> [(Dice 1, Dice 2)]. We need to consider both the cases here.

Total possibilities are 6 * 6 = 36.

So, probability(sum=3) = 2 / 36 = 1/18
According to me, this example is quite similar to the given prob. My understanding might be wrong as well.

Experts, please throw some light here..!
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by papgust » Thu Nov 26, 2009 9:37 pm
Perminology,

I was really confused and my mind jumped from one answer to another. I sat down for a few minutes and finally i have realized why B is the correct answer

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

These are the possible combinations --> (2,7), (3,6), (4,5), (5,4)

Should we consider the combinations (6,3) (7,2)? Answer is NO!

We have already selected 3 from A and 6 from B. For (6,3), we are again selecting 3 from A and 6 from B [Because A doesn't have 6 and B doesn't have 3]

This applies to (7,2) as well.

So, the right answer is B (0.20).

-------------

Let's change the sets A and B slightly,
A=(2,3,4,5,6)
B=(2,3,4,5,6)
Now, Let's find the combinations satisfying the same restriction (Sum=9),

These are the possible combinations --> (3,6), (6,3), (4,5), (5,4)

Here it's a different case. We need to consider both (3,6) and (6,3) because both A and B have 3 and 6. This applies to (4,5) and (5,4) as well.

I hope this clears your doubt..!
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by salar_snake » Thu Nov 26, 2009 11:36 pm
This problem isn't that complicated, I mean one number is chosen from each set, right ? and we know that there are four combinations that add up to nine:

(2,7), (3,6), (4,5), (5,4) [the first number is from set A and the second number is from set B]

Does it matter which number we choose from set A ? no, you can choose any number, for instance, take 2.

Now the probability of choosing the right number from set B is : 1/5th = 0.2 (number 7 has to be chosen form set B)
for any other number its the same.
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