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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 `x_i` be the number on the card drawn from the ith box, i = 1, 2, 3. The probability that `x_1, x_2, x_3` are in an aritmetic progression is

A

`(9)/(105)`

B

`(10)/(105)`

C

`(11)/(105)`

D

`(7)/(105)`

Text Solution

Verified by Experts

The correct Answer is:
C
Here, `2x_(2) = x_(1) + x_(3)`
`implies x_(1) + x_(3)` = even
So, either `x_(1)` and `x_(3)` are both odd or both even.
Hence, number of favorable ways = `.^(2)C_(1).^(4)C_(1) + .^(1)C_(1) .^(3)C_(1) = 11`.
Therefore, required probability is `(11)/(105)`.
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