It has been found that if $$A$$ and $$B$$ play a game $$12$$ times, $$A$$ wins $$6$$ times, $$B$$ wins $$4$$ times and they draw twice. $$A$$ and $$B$$ take part in a series of $$3$$ games. The probability that they will win alternately is :
A.
$$\frac{5}{{72}}$$
B.
$$\frac{5}{{36}}$$
C.
$$\frac{{19}}{{27}}$$
D.
none of these
Answer :
$$\frac{5}{{36}}$$
Solution :
The probability of $$A$$ winning in a game $$ = P\left( A \right) = \frac{6}{{12}} = \frac{1}{2}$$
The probability of $$B$$ winning in a game $$ = P\left( B \right) = \frac{4}{{12}} = \frac{1}{3}$$
The required probability
$$\eqalign{
& = P\left( {A \cap B \cap A} \right) + P\left( {B \cap A \cap B} \right) \cr
& = P\left( A \right).P\left( B \right).P\left( A \right) + P\left( B \right).P\left( A \right).P\left( B \right) \cr
& = \frac{1}{2}.\frac{1}{3}.\frac{1}{2} + \frac{1}{3}.\frac{1}{2}.\frac{1}{3} \cr
& = \frac{5}{{36}}. \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$