Rahul has to write a project, Probability that he will get a project copy is $$'p'$$, probability that he will get a blue pen is $$'q'$$ and probability that he will get a black pen is $$\frac{1}{2}$$. If he can complete the project either with blue or with black pen or with both and probability that he completed the project is $$\frac{1}{2}$$ then $$p\left( {1 + q} \right)$$ is :
A.
$$\frac{1}{2}$$
B.
$$1$$
C.
$$\frac{1}{4}$$
D.
$$2$$
Answer :
$$1$$
Solution :
Lets define the events as
Probability of getting project copy $$\left( A \right) = p$$
Probability of getting blue pen $$\left( B \right) = q$$
Probability of getting black pen $$\left( C \right) = \frac{1}{2}$$
Then, $$\eqalign{
& {\text{ }}p\left( {AB\overline C } \right) + p\left( {AC\overline B } \right) + p\left( {ABC} \right) = \frac{1}{2} \cr
& p.q.\frac{1}{2} + p.\frac{1}{2}\left( {1 - q} \right) + p.q.\frac{1}{2} = \frac{1}{2} \cr
& \therefore \,pq + p - pq + pq = 1 \cr
& \therefore \,p\left( {1 + q} \right) = 1 \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$