Three randomly chosen non-negative integers x | vedclass

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(B) The total number of non-negative integer solutions to $x+y+z=n$ is given by the formula $\binom{n+k-1}{k-1}$,where $k=3$ and $n=10$.Total solution

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$\frac{6}{11}$

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(B) The total number of non-negative integer solutions to $x+y+z=n$ is given by the formula $\binom{n+k-1}{k-1}$,where $k=3$ and $n=10$.Total solutions $= \binom{10+3-1}{3-1} = \binom{12}{2} = \frac{12 	imes 11}{2} = 66$.For $z$ to be even,let $z \in \{0, 2, 4, 6, 8, 10\}$.If $z=k$,then $x+y=10-k$. The number of solutions for $x+y=m$ is $m+1$.For $z=0$,$x+y=10$,solutions $= 11$.For $z=2$,$x+y=8$,solutions $= 9$.For $z=4$,$x+y=6$,solutions $= 7$.For $z=6$,$x+y=4$,solutions $= 5$.For $z=8$,$x+y=2$,solutions $= 3$.For $z=10$,$x+y=0$,solutions $= 1$.Total favourable solutions $= 11+9+7+5+3+1 = 36$.Probability $P = \frac{36}{66} = \frac{6}{11}$.

Three randomly chosen non-negative integers $x, y$ and $z$ are found to satisfy the equation $x+y+z=10$. Then the probability that $z$ is even,is

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