Question

A bag contains $4$ red balls and $6$ black balls. A ball is drawn at random from the bag, its color is observed and this ball along with two additional balls of the same color 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. $\dfrac{1}{5}$
B. $\dfrac{3}{4}$
C. $\dfrac{3}{10}$
D. $\dfrac{2}{5}$

Answer 1

Hint: We use the concepts of conditional probability to solve questions of probability where the probability of happening or not happening of an event depends on another event. Firstly, we need to apply the formula of conditional probability. Then, substitute the values and simplify to get the answer.

Complete step by step solution:
We are given a bag containing $4$ red balls and $6$ black balls. A ball is drawn at random from the bag, its color is observed and this ball along with two additional balls of the same color are returned to the bag.
From the above, we can see that there can be two possible cases in this situation, the first case involves drawing a red ball and the second case involves drawing a black ball.
Finding out the probability of drawing the second ball as a red ball when the first ball drawn is black, we get,
$\Rightarrow \dfrac{4}{4+8}=\dfrac{1}{3}$
Finding out the probability of drawing the second ball as a red ball when the first ball drawn is red, we get,
$\Rightarrow \dfrac{4+2}{6+6}=\dfrac{1}{2}$
From the above, the probability of the second ball being red can be given as follows,
Probability of first ball being black $\times$ Probability of second ball being red when the first ball is black $+$ probability of first ball being red $\times$ Probability of second ball being red when the first ball is red.
$\Rightarrow \dfrac{6}{10}\times \dfrac{1}{3}+\dfrac{4}{10}\times \dfrac{1}{2}$
Simplifying the above expression, we get,
$\Rightarrow \dfrac{6}{30} + \dfrac{4}{20}$
Cancelling out the common factors, we get,
$\Rightarrow \dfrac{2}{5}$

Therefore, the probability of drawing a red ball is $\dfrac{2}{5}$.

Note: We often tend to confuse the conditional probability of one event with the other and fail to recognize when the sample space changed while solving questions with the conditional probability concept being used. We also confuse conditional events for sequential events which leads to application of the wrong formula.