SATHEE: Probability 1 Question 6

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) $\frac{36}{55}$

(d) $\frac{5}{11}$

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Answer:

Correct Answer: 6. (c)

Solution:

Sample space $\to^{12} C_{2}$

Number of possibilities for $z$ is even.

$\begin{aligned} z = 0 \Rightarrow^{11} C_{1} \\ z = 2 \Rightarrow^{9} C_{1} \\ z = 4 \Rightarrow^{7} C_{1} \\ z = 6 \Rightarrow^{5} C_{1} \\ z = 8 \Rightarrow^{3} C_{1} \\ z = 10 \Rightarrow^{1} C_{1} \end{aligned}$

$Total = 36$

$∴$ Probability $= \frac{36}{66} = \frac{6}{11}$

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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

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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