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Question

Box 1 contains three cards bearing numbers 1,2,3; box 2 contains five cards bearing numbers 1,2,3,4,5; and box 3 contains seven cards bearing numbers 1,2,3,4,5,6,7. A card is drawn from each of the boxes. Let xi be the number on the card drawn from the ith box, i=1,2,3.

The probability that x1,x2,x3 are in an arithmetic progression, is

A
9105
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B
10105
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C
11105
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D
7105
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Solution

The correct option is C 11105
x1,x2,x3 are in A.P.
x1+x3=2x2
x1+x3=even
x1,x3 both should be either even or odd
So, total number of favourable cases is 1c1.3c1+2c1.4c1=11.
Required probability =113×5×7=11105

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