JEE Advanced 2017 · Paper 2 · Mathematics Q7

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MathematicsSingle correct+3 / -1Question 43 of 54

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

\frac{36}{55}

\frac{6}{11}

\frac{1}{2}

\frac{5}{11}

Answer and worked solution

Official answer: B

Total. Nonnegative solutions of $x+y+z=10$ : $\binom{12}2=66$ .

z even. For $z=2m$ , $x+y=10-2m$ has $11-2m$ solutions. Summing over $m=0,\dots,5$ :
$11+9+7+5+3+1=36.$

Probability. $\dfrac{36}{66}=\dfrac6{11}$ .

(B)

Topics:Classical probability Distribution and arrangements

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JEE Advanced 2017 · Paper 2, Mathematics question 7