Three randomly chosen nonnegative integers x, y and z are found to satisfy equation x+y+z=10 . Then

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14.Probability

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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

$\frac{36}{55}$

$\frac{6}{11}$

$\frac{1}{2}$

$\frac{5}{11}$

(IIT-2017)

Solution

Total number of solutions $={ }^{10+3-1} \mathrm{C}_{3-1}=66$

Favourable number of solutions $={ }^{11} \mathrm{C}_1+{ }^9 \mathrm{C}_1+{ }^7 \mathrm{C}_1+{ }^5 \mathrm{C}_1+{ }^3 \mathrm{C}_1+{ }^1 \mathrm{C}_1=36$

$\mathrm{P}(\mathrm{req})=\frac{36}{66}=\frac{6}{11}$

Standard 11

Mathematics

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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

Solution

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