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Three randomly chosen nonnegative intege...

Three randomly chosen nonnegative integers `x , ya n dz` are found to satisfy the equation `x+y+z=10.` Then the probability that `z` is even, is: `5/(12)` (b) `1/2` (c) `6/(11)` (d) `(36)/(55)`

A

`(1)/(2)`

B

`(36)/(55)`

C

`(6)/(11)`

D

`(5)/(11)`

Text Solution

Verified by Experts

The correct Answer is:
C
`x + y + z = 10`
Total number of non-negative solutions = `.^(10+3-1)C_(3-1) = .^(12)C_(2) = 66`
Now let z = 2n
`therefore x + y = 10 - 2n, n ge 0`
Number of non-negative solutions of above equation is
11 + 9 + 7 + 5 + 3 + 1 = 36.
`therefore` Required probability = `(36)/(66) = (6)/(11)`
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