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probability quest 1

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probability quest 1

by engg.manik » Sun Oct 04, 2009 9:19 am
Q-> There are two set each with the number 1, 2, 3, 4, 5, 6. If randomly choose one number from each set, what is the probability that the product of the 2 numbers is divisible by 4?
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by mehravikas » Sun Oct 04, 2009 3:08 pm
Is it 15/36?
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by sanjana » Mon Oct 05, 2009 9:39 am
IMO : 13/36

Picking 2 numbers from each set : 6c1*6c1=36

Favourable outcomes = 13
(1,4)
(2,4),(2,6)
(3,4)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,4)
(6,2),(6,4)

Therefore Required prob = 13/36.

Whats the OA?
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by mehravikas » Mon Oct 05, 2009 11:58 am
What about (2,2) and (6, 6)
sanjana wrote:IMO : 13/36

Picking 2 numbers from each set : 6c1*6c1=36

Favourable outcomes = 13
(1,4)
(2,4),(2,6)
(3,4)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,4)
(6,2),(6,4)

Therefore Required prob = 13/36.

Whats the OA?
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by engg.manik » Mon Oct 05, 2009 5:22 pm
OA is 5/12
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by engg.manik » Mon Oct 05, 2009 5:23 pm
OA is 5/12
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by sanjana » Mon Oct 05, 2009 5:46 pm
Yes missed (2,2) and (6,6) hence the answer is 5/12.

Thanks!
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by tapdoanhp » Sat Mar 03, 2012 3:21 am
I have another solution to this prob without listing all possible cases.

+ If the element taken out from the first set is 4 then the left is what ever element from set 2
Prob=1/6
+ If the first element is even then the left element have to be even
Prob=2/6 (eliminate 4)*3/6=1/6
+ If the element is odd then the left must be 4
Prob=3/6*1/6=1/12

Total Prob = 1/6+1/6+1/12 =5/12
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