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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 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` is odd is The probability that `x_1, x_2, x_3` are in an aritmetic progression is

A

`(29)/(105)`

B

`(53)/(105)`

C

`(57)/(105)`

D

`(1)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
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)`
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