A bag contains 4 red and 3 black balls. A second bag contains 2 red and 3 black balls. One bag is selected…
A bag contains 4 red and 3 black balls. A second bag contains 2 red and 3 black balls. One bag is selected at random. If from the selected bag, one ball is drawn at random, then the probability that the ball drawn is red, is
$\frac{39}{70}$
$\frac{41}{70}$
$\frac{29}{70}$
$\frac{17}{35}$
Solution
A red ball can be drawn in two mutually exclusive ways.
(i) Selecting bag I and then drawing a red ball from it.
(ii) Selecting bag II and then drawing a red ball from it
Let $E_1^{\prime}, E_2$ and $A$ denote the events defined as follows
$\begin{aligned} & E_1=\text { Selecting bag I } \\ & E_2=\text { Selecting bag II }\end{aligned}$
Since, one of the two bags is selected randomly.
$\therefore P\left(E_1\right)=\frac{1}{2}$ and $P\left(E_2\right)=\frac{1}{2}$
Now, $P\left(\frac{A}{E_1}\right)=$ Probability of drawing a red ball
when the first bag has been selected $=4 / 7$
$P\left(\frac{A}{E_2}\right)=$ Probability of drawing a red ball when
the second bag has been selected $=2 / 5$
$P($ red ball $)=P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)$
$=\frac{1}{2} \times \frac{4}{7}+\frac{1}{2} \times \frac{2}{5}=\frac{2}{7}+\frac{1}{5}=\frac{17}{35}$