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Question

There are two die A and B both having six faces. Die A has three faces marked with 1, two faces marked with 2, and one face marked with 3. Die B has one face marked with 1, two faces marked with 2, and three faces marked with 3. Both dices are thrown randomly once. If E be the event of getting sum of the numbers appearing on top faces equal to x, let P(E) be the probability of event E, then
When x=4, then P(E) is equal to

A
59
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B
67
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C
718
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D
819
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Solution

The correct option is C 718
x can be 2,3,4,5,6. The number of ways in which sum of 2,3,4,5,6 can occur by the coefficients of x2,x3,x4,x4,x5,x6 in
(3x+2x2+x3)(x+2x2+3x3)=3x2+8x3+14x4+8x5+3x6
The number of ways in which different sums can occur is
(3+2+1)(1+2+3)=36.
The probability of 4 is 1436=718

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