Question

A bag contains $4$ red and $6$ black balls. A ball is drawn at random from the bag and its colour is observed and this ball along with additional two balls of the same colour are returned to the bag. If now a ball is drawn at random from this bag, then the probability that this ball drawn is red is:
A. $\dfrac{1}{5}$
B. $\dfrac{3}{4}$
C. $\dfrac{3}{{10}}$
D. $\dfrac{2}{5}$

Answer 1

At first, make two cases where in the first case, we draw black balls at the first and the second where we draw the red ball. Find their respective probabilities. Then add two balls for each case and find the probability of the red ball.

Complete step-by-step answer:
Let us divide the problem into two cases;
Case $1$: when we draw black ball at first and now we will find the probability of the black ball. But why so?
This is because the probability of finding the red ball second time depends on what we take out first time.
$\Rightarrow$ $P({\text{black ball) = }}\dfrac{6}{{4 + 6}} = \dfrac{3}{5}$ $- - - - (1)$
As stated in the problem, we will add two more balls of the colour we took which means that we will add two more black balls.
Total number of black balls now $= 8$
Now probability of getting the red ball $= \dfrac{4}{{4 + 8}} = \dfrac{1}{3}$ $- - - - - (2)$
Case $1$ is now completed.
But now the way to find the total probability of getting the red ball from the first case is by multiplying (1) and (2) because one leads to another.
$\Rightarrow$ $P({\text{red ball) = }}\dfrac{3}{5} \times \dfrac{1}{3} = \dfrac{1}{5}$
Similarly we will make the second case:
$\Rightarrow$ $P({\text{red ball) = }}\dfrac{4}{{4 + 6}} = \dfrac{2}{5}$
Now we will add two more red balls to it and then we will get the total of six red balls.
$\Rightarrow$ $P({\text{red ball) = }}\dfrac{6}{{6 + 6}} = \dfrac{1}{2}$
Therefore from the second case,
$\Rightarrow$ $P({\text{red ball) = }}\dfrac{2}{5} \times \dfrac{1}{2} = \dfrac{1}{5}$
Now we will add both the cases
$\Rightarrow$ $P({\text{red ball) = }}\dfrac{1}{5} + \dfrac{1}{5} = \dfrac{2}{5}$

Hence option D is the correct answer.

Note: Sum of Probability of an event and the probability of that event to not occur is always equal to 1. If P(E) be the probability of the event to occur and P(E’) be the probability of the event to not occur, then
$P(E)+P(E’)=1$.