Amar, Bimal and Chetan are three contestants for an election, odds against Amar will win is $$4 : 1$$ and odds against Bimal will win is $$5 : 1$$ and odds in favor of Chetan will win $$2 : 3$$ then what is probability that either Amar or Bimal or Chetan will win the election :
A.
$$\frac{{23}}{{20}}$$
B.
$$\frac{{11}}{{30}}$$
C.
$$\frac{7}{{10}}$$
D.
none of these
Answer :
none of these
Solution :
From the given information probability that
Amar will win the tournament is $$P\left( A \right) = \frac{1}{5}$$ and Bimal will win $$P\left( B \right) = \frac{1}{6}$$ and same for Chetan is $$P\left( C \right) = \frac{2}{5}.$$
Since these events are mutually exclusive.
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$