Probability- Multiplication and additional theorem question

Question

In a urn there are 3 red and 7 white balls, one ball is choose at random, then selected ball is replaced with the ball of another colour, now the ball is selected randomly what is the probability is that selected ball is red?

Now I have worked on it Queries

one of faculty is saying "another ball means it can be any another colour ball like pink or black etc"

and my view is that in first selection if we get red then new white ball would be added, as in sample space we have only red and white and another ball means not red

Provide your view on it

in my case probability would be 34/100

Answer 1

Hello Varun,

My view on this is that the question may be a little ambiguous. The line "selected ball is replaced with the ball of another colour" could be interpreted as any of the following scenarios.

In each scenario, we can calculate the desired probability [P(2nd ball red)] by considering the different ways it can occur and their conditional probabilities:

P(2nd ball red) = P(2nd ball red | 1st ball red) * P(1st ball red) + P(2nd ball red | 1st ball white) * P(1st ball white)

1. Scenario #1 (the most straightforward, in my opinion): "the ball of another colour" means if a white ball is drawn then it is replaced with a red ball, or if a red ball is drawn it is replaced by a white ball.

This scenario gives the answer you came up with: 34/100.

P(2nd ball red) = P(2nd ball red | 1st ball red) * P(1st ball red) + P(2nd ball red | 1st ball white) * P(1st ball white)

= 2/10 * 3/10 + 4/10 * 7/10 = 34/100

1. Scenario #2: "the ball of another colour" means a non-white and non-red ball (e.g. pink, black, blue) is added to the urn after the first ball is drawn (red or white).

Then, P(2nd ball red) = P(2nd ball red | 1st ball red) * P(1st ball red) + P(2nd ball red | 1st ball white) * P(1st ball white)

= 2/10 * 3/10 + 3/10 * 7/10 = 27/100

1. Scenario #3: "the ball of another colour" means if a red ball is drawn, it is replaced with a ball of any other color (including white) or if a white ball is drawn, it is replaced with a non-white ball (e.g. pink, black, blue, red).

This scenario is impossible to solve without more information or making assumptions about the probability of replacing a drawn white ball with a red ball.

Suppose the probability of replacing a drawn white ball with a red ball is p.

Then, P(2nd ball red) = P(2nd ball red | 1st ball red) * P(1st ball red) + P(2nd ball red | 1st ball white & replaced by red) * P(1st ball white) + P(2nd ball red | 1st ball white & replaced by non-red) * P(1st ball white)

= 2/10 * 3/10 + 4/10 * p * 7/10 + 3/10 * (1-p) * 7/10 = (27 + 7p)/100

**Note that Scenarios 1 and 2 are special cases of Scenario 3 in which p = 1 or 0, respectively.

I hope this is helpful.

-Josh