An ordinary dice is rolled a certain number of times. The probability of getting an odd number $$2$$ times is equal to the probability of getting an even number $$3$$ times. Then the probability of getting an odd number an odd number of times is :
A.
$$\frac{1}{{32}}$$
B.
$$\frac{5}{{16}}$$
C.
$$\frac{1}{2}$$
D.
none of these
Answer :
$$\frac{1}{2}$$
Solution :
The probability of getting an odd number in a throw $$ = \frac{3}{6} = \frac{1}{2}$$
The probability of getting an odd number $$2$$ times in $$n$$ trials $$ = {}^n{C_2}.{\left( {\frac{1}{2}} \right)^2}.{\left( {\frac{1}{2}} \right)^{n - 2}}$$
Similarly, the probability of getting an even number $$3$$ times in $$n$$ trials $$ = {}^n{C_3}.{\left( {\frac{1}{2}} \right)^3}.{\left( {\frac{1}{2}} \right)^{n - 3}}$$
From the question, $$ {}^n{C_2}.{\left( {\frac{1}{2}} \right)^n} = {}^n{C_3}.{\left( {\frac{1}{2}} \right)^n}\,\,\,\,\, \Rightarrow n = 5$$
$$\therefore $$ the required probability
$$\eqalign{
& = {}^5{C_1}.\left( {\frac{1}{2}} \right).{\left( {\frac{1}{2}} \right)^4} + {}^5{C_3}.{\left( {\frac{1}{2}} \right)^3}.{\left( {\frac{1}{2}} \right)^2} + {}^5{C_5}.{\left( {\frac{1}{2}} \right)^5} \cr
& = \left( {5 + 10 + 1} \right).\frac{1}{{{2^5}}} \cr
& = \frac{{16}}{{32}} \cr
& = \frac{1}{2} \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$