A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is:
A.
$$\frac{2}{5}$$
B.
$$\frac{1}{5}$$
C.
$$\frac{3}{4}$$
D.
$$\frac{3}{10}$$
Answer :
$$\frac{2}{5}$$
Solution :
Let $${R_t}$$ be the even of drawing red ball in $${t^{th}}$$ draw and $${B_t}$$ be the event of drawing black ball in $${t^{th}}$$ draw.
Now, in the given bag there are 4 red and 6 black balls.
$$\eqalign{
& \therefore \,\,P\left( {{R_1}} \right) = \frac{4}{{10}}\,{\text{and }}P\left( {{B_1}} \right) = \frac{6}{{10}} \cr
& {\text{And, }}P\left( {\frac{{{R_2}}}{{{R_1}}}} \right) = \frac{6}{{12}}\,{\text{and }}P\left( {\frac{{{R_2}}}{{{B_1}}}} \right) = \frac{4}{{12}} \cr} $$
Now, required probability
$$\eqalign{
& = P\left( {{R_1}} \right) \times P\left( {\frac{{{R_2}}}{{{R_1}}}} \right) + P\left( {{B_1}} \right) \times P\left( {\frac{{{R_2}}}{{{B_1}}}} \right) \cr
& = \left( {\frac{4}{{10}} \times \frac{6}{{12}}} \right) + \left( {\frac{6}{{10}} \times \frac{4}{{12}}} \right) \cr
& = \frac{2}{5} \cr} $$
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.
$$\frac{{P\left( {\overline A } \right)}}{{P\left( {\overline B } \right)}}$$