Probability 1 Question 6
6. 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
(a) $\frac{1}{2}$
(b) $\frac{36}{55}$
(c) $\frac{6}{11}$
(d) $\frac{5}{11}$
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Answer:
Correct Answer: 6. (c)
Solution:
- Sample space $\rightarrow{ }^{12} C _2$
Number of possibilities for $z$ is even.
$$ \begin{aligned} & z=0 \Rightarrow{ }^{11} C _1 \\ & z=2 \Rightarrow{ }^{9} C _1 \\ & z=4 \Rightarrow{ }^{7} C _1 \\ & z=6 \Rightarrow{ }^{5} C _1 \\ & z=8 \Rightarrow{ }^{3} C _1 \\ & z=10 \Rightarrow{ }^{1} C _1 \end{aligned} $$
$$ \text { Total }=36 $$
$\therefore$ Probability $=\frac{36}{66}=\frac{6}{11}$
