JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Permutations and Combinations

Three randomly chosen nonnegative integers

x, y

and

z

are found to satisfy the equation

x + y + z = 10

. Then the probability that

z

is even, is

Answer: (B)

Solution

The total number of nonnegative integer solutions to

x + y + z = 10

is:

(3 - 1 10 + 3 - 1) = (2 12) = 66

For

z

to be even, let

z = 2 m

where

m \geq 0

. Then

x + y = 10 - 2 m

.

For each value of

m

from 0 to 5:

$m = 0$ : $z = 0$ , $x + y = 10$ → $(1 11) = 11$ solutions
$m = 1$ : $z = 2$ , $x + y = 8$ → $(1 9) = 9$ solutions
$m = 2$ : $z = 4$ , $x + y = 6$ → $(1 7) = 7$ solutions
$m = 3$ : $z = 6$ , $x + y = 4$ → $(1 5) = 5$ solutions
$m = 4$ : $z = 8$ , $x + y = 2$ → $(1 3) = 3$ solutions
$m = 5$ : $z = 10$ , $x + y = 0$ → $(1 1) = 1$ solution

Total favorable outcomes:

11 + 9 + 7 + 5 + 3 + 1 = 36

Probability:

\frac{36}{66} = \frac{6}{11}

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option