Question

A bag contains 2 red balls and 2 green balls. A person randomly pulls out a ball, replacing it with a red ball regardless of the colour. What is the probability that all the balls are red after three such replacements?
A. $\dfrac{3}{8}$
B. $\dfrac{7}{16}$
C. $\dfrac{5}{32}$
D. $\dfrac{9}{32}$

Answer 1

Here we have been given a bag in which there are 2 red balls and 2 green balls. We have then been told that a person picks up a ball at random and replaces it with a red ball irrespective of its colour and this process is repeated 3 times and we have to find the probability that after that, all the balls in the bag are red. For this, we will first see that both the green balls should be removed for that to happen and the third ball would of course be a red ball which wouldn’t make a difference. So, we will find the probability of the fact that 2 of the picked balls are green out of the 3 balls. We will make different cases in which order all the balls can be picked and then find the probability of all the cases and then we will add the total probability and hence we will get the required answer.

Complete step by step answer:
Here the bag contains 2 red and 2 green balls. We have been told that a random ball is being picked from the bad and is being replaced by a red ball irrespective of its colour. Thus, doesn’t matter if the ball is green or red, it will still be replaced by another ball of red colour.
Now, we have also been told that this process is repeated 3 times.
Now, for all the balls to be red after the process is completed, the person should pick both the green balls and then of course only red balls are remaining, so the third ball that has to be picked will be red.
Now, the balls can be picked in the following order:
1. RGG
2. GRG
3. GGR
Here, R represents the red coloured ball and G the green coloured ball.
Now, if red ball is being picked and replaced, it will not make any change and we will count the probability here of picking red balls as 1. Hence, the probability that all the balls are green after this process will be calculated only through the probability of picking of the green balls.
Now, we will find the probability of each case.
Case-1: RGG
Here, there are a total of 4 balls out of 2 are red. Now, there are still 2 green balls left. Thus, the probability that the second ball picked is green is given as:
$1\times \dfrac{2}{4}$
Now, there is only 1 green ball left and still the total number of balls in the bag is 4. Hence, the probability of the case RGG is:
$\begin{aligned} & 1\times \dfrac{2}{4}\times \dfrac{1}{4} \\\ & \Rightarrow \dfrac{1}{8} \\\ \end{aligned}$
Case-2: GRG
Here, there are a total of 4 balls and first ball being picked is green and the second being picked is red. Thus the probability of this happening is given as:
$\dfrac{2}{4}\times 1$
Now, there is again only one green ball remaining and still there are a total of 4 balls in the bag. Hence, the probability of the case GRG is given as:
$\begin{aligned} & \dfrac{2}{4}\times 1\times \dfrac{1}{4} \\\ & \Rightarrow \dfrac{1}{8} \\\ \end{aligned}$
Case-3: GGR
Here, the first ball being picked is green, the probability of this happening is given as:
$\dfrac{2}{4}$
Now, the second ball is also green but now there is only 1 green ball left out of the 4 and the last ball being picked is red only. Hence, the probability of GGR case is given as:
$\begin{aligned} & \dfrac{2}{4}\times \dfrac{1}{4}\times 1 \\\ & \Rightarrow \dfrac{1}{8} \\\ \end{aligned}$
Now, the total probability that all the balls are red after replacing any three balls at random is given as:
$\begin{aligned} & \dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8} \\\ & \Rightarrow \dfrac{3}{8} \\\ \end{aligned}$
Hence, the required probability is $\dfrac{3}{8}$.

So, the correct answer is “Option A”.

Note: Probability follows the same order as does the questions in permutations and combinations. If two or more events have to happen simultaneously, then the total probability of those combined events is equal to the probability of the individual events and if any one of the multiple events have to happen, then the total probability of the total even is equal to the sum of those individual events. Here, both these cases have been used simultaneously.