Rajesh doesn’t like to study. Probability that he will study is $$\frac{1}{3}.$$ If he studied then probability that he will fail is $$\frac{1}{2}$$ and if he didn’t study then probability that he will fail is $$\frac{3}{4}.$$ If in result Rajesh failed then what is the probability that he didn’t studied.
A.
$$\frac{2}{3}$$
B.
$$\frac{3}{4}$$
C.
$$\frac{1}{3}$$
D.
none of these
Answer :
$$\frac{3}{4}$$
Solution :
He will fail in exam in two cases :
$${\bf{Case}}\,\left( {\bf{i}} \right)\,{\bf{:}}$$ He studied and failed, probability of this case is $$\left( {\frac{1}{3}} \right)\left( {\frac{1}{2}} \right) = \frac{1}{6}$$
$${\bf{Case}}\,\left( {{\bf{ii}}} \right)\,{\bf{:}}$$ He didn’t studied and failed, probability of this case is $$\left( {\frac{2}{3}} \right)\left( {\frac{3}{4}} \right) = \frac{1}{2}$$
So total probability is $$\frac{1}{6} + \frac{1}{2} = \frac{4}{6} = \frac{2}{3}$$
Then required probability $$ = \frac{{\left( {\frac{1}{2}} \right)}}{{\left( {\frac{2}{3}} \right)}} = \frac{3}{4}$$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$