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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
Plan `"Probability "=("Number of favourable outcomes ")/("Number of total outcomes ")`
As `x_1 +x_2+x+3 ` is odd.
So, all may be odd or one of them is odd and other two are even.
`therefore` Required probability
`(""^(2)C_1xx""^(3)C_1xx""^(4)C_1+""^(1)C_1xx""^(2)C_1xx""^(4)C_1+""^(2)C_1xx""^(2)C_1xx""^(3)C_1+""^(1)C_1xx""^(3)C_1xx""^(7)C_1)/(""^3C_1xx""^(5)C_1xx""^(7)C_1)`
`(24+8+12+9)/(105)`
`(53)/(105)`
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