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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

`(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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IIT JEE PREVIOUS YEAR-PROBABILITY-TOPIC 1 CLASSICAL PROBABLITY OBJECTIVE QUESTIONS I (PASSAGE BASED PROBLEMS )
  1. Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains fiv...

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