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Box 1 contains three cards bearing numbe...

Box 1 contains three cards bearing numbers, 1, 2, 3, box 2 contains five cards bearing number 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 `i^(th)` box, i = 1, 2, 3.
The probability that `x_(1), x_(2), x_(3)` are in the arithmetic progression, is

A

`(9)/(105)`

B

`(10)/(105)`

C

`(11)/(105)`

D

`(7)/(105)`

Text Solution

Verified by Experts

The correct Answer is:
C
Since `x_1,x_2,x_3` are in AP.
`therefore x_1+x_3=2x_2`
So `x_1+x_3` should be even number
Either both `x_1 and x_3` are odd or both are even.
`therefore` Required probability `=(""^(2)C_1xx""^(4)C_1+""^(1)C_1xx""^(3)C_1)/(""^(3)C_1xx""^(5)C_1xx""^(7)C_1)`
`=11/105`
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