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Question

Box I contain three cards bearing numbers 1,2,3; box II contains five cards bearing numbers 1,2,3,4,5; and box III 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 and x3 are in 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


Since, x1,x2,x3 are in AP.
x1+x3=2x2
So,x1+x3 should be even number.
Either both x1 and x3 are odd or both are even.
Required probability=2C1×4C1+1C1×3C13C1×5C1×7C1 =11105


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