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.
$$\frac{{36}}{{55}}$$
B.
$$\frac{{6}}{{11}}$$
C.
$$\frac{{1}}{{2}}$$
D.
$$\frac{{5}}{{11}}$$
Answer :
$$\frac{{6}}{{11}}$$
Solution :
Total number of non negative solutions of $$x + y + z = 10$$ are $$^{12}{C_2} = 66$$ (using $$^{n + r - 1}{C_{r - 1}}$$ )
If $$z$$ is even then there can be following cases :
$$z = 0$$
⇒ No. of ways of solving $$x + y = 10$$
⇒ $$^{11}{C_1}$$
$$z = 2$$
⇒ No. of ways of solving $$x + y = 8$$
⇒ $$^{9}{C_1}$$
$$z = 4$$
⇒ No. of ways of solving $$x + y = 6$$
⇒ $$^{7}{C_1}$$
$$z = 6$$
⇒ No. of ways of solving $$x + y = 4$$
⇒ $$^{5}{C_1}$$
$$z = 8$$
⇒ No. of ways of solving $$x + y = 2$$
⇒ $$^{3}{C_1}$$
$$z = 10$$
⇒ No. of ways of solving $$x + y = 0$$
⇒ 1
∴ Total ways when $$z$$ is even = 11 + 9 + 7 + 5 + 3 + 1 = 36
∴ Required probability $$ = \frac{{36}}{{66}} = \frac{6}{{11}}$$
Releted MCQ Question on Statistics and Probability >> Probability
Releted Question 1
Two fair dice are tossed. Let $$x$$ be the event that the first die shows an even number and $$y$$ be the event that the second die shows an odd number. The two events $$x$$ and $$y$$ are:
Two events $$A$$ and $$B$$ have probabilities 0.25 and 0.50 respectively. The probability that both $$A$$ and $$B$$ occur simultaneously is 0.14. Then the probability that neither $$A$$ nor $$B$$ occurs is
The probability that an event $$A$$ happens in one trial of an experiment is 0.4. Three independent trials of the experiment are performed. The probability that the event $$A$$ happens at least once is
If $$A$$ and $$B$$ are two events such that $$P(A) > 0,$$ and $$P\left( B \right) \ne 1,$$ then $$P\left( {\frac{{\overline A }}{{\overline B }}} \right)$$ is equal to
(Here $$\overline A$$ and $$\overline B$$ are complements of $$A$$ and $$B$$ respectively).
A.
$$1 - P\left( {\frac{A}{B}} \right)$$
B.
$$1 - P\left( {\frac{{\overline A }}{B}} \right)$$
C.
$$\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\overline B } \right)}}$$
D.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$