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

Options

A. `(29)/(105)`
B. `(53)/(105)`
C. `(57)/(105)`
D. `(1)/(2)`

Answer

B

Explanation

Fo, `x_(1) + x_(2) + x_(3)` = odd.
Case I: One odd, two even
(OEE) or (EOE) or (EEO)
Total number of ways = `2xx2xx3+1xx3xx3+1xx2xx4=29`.
Case II: All three odd
Number of ways = `2xx3xx4=24`
`therefore` Favorable ways = 53
`therefore` Required probability = `(53)/(3xx5xx7) = (53)/(105)`