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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probability quest 1
Source: Beat The GMAT — Problem Solving |
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?
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?
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?
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
+ 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
