 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says An art contains five red and five black balls. A ball is drawn at random Its color is noted and is returned to the urn Moreover two additional balls of the color drawn are put in the urn And then a ball is drawn at random. What is the probability that the second ball is red Let us first understand the theorem of total probability Let even e2 so on en be a partition of the sample space S That is they are pair-wise disjoint exhaustive and have non-zero probabilities and Suppose that each of the events Even e2 so on en has non-zero probability of occurrence Let a be any event Associated with S then probability of a is equal to probability of even Into probability of a upon even plus Probability of e2 Into probability of a upon e2 plus so on Probability of e n into probability of a upon En So this is a key idea behind our question We will take the help of this key idea to solve the question So let's start the solution Now according to the question and on contains five red and five black balls a Ball is drawn at random. Its color is noted and is returned to the urn let even be the event That the first ball drawn is red e2 be the event That the first ball drawn is black therefore probability of e1 is equal to five upon ten and this is equal to one upon two and The probability of e2 is Equal to five upon ten which is again equal to one upon two Now again according to the question two additional Balls of the color drawn are put in the urn and then a ball is drawn at random We have to find the probability that the second ball is red Let a be the event that The second ball is Red Now the conditional probability that the second ball is red given first ball drawn is also red that is probability of a upon e1 is equal to seven upon twelve Because two additional balls of the color drawn are put in the urn So total red balls are now five plus two which is equal to seven Hence the probability of a upon e1 is equal to seven upon twelve Again conditional probability That the second ball is red given first ball drawn is black that is probability of a upon e2 and This is equal to five upon twelve as Total red balls are still five now events even and e2 are pairwise disjoint Exhausted and have non-zero probabilities So they represent a partition of the sample space as since events Even and e2 forms a Partition of the sample space as therefore by Theorem on total probability we have probability of a is equal to probability of e1 into probability of a upon e1 plus probability of e2 into probability of a upon e2 now we have probability of e1 is equal to probability of e2 which is equal to one upon two and Probability of a upon e1 is seven upon twelve and the probability of a upon e2 is five upon twelve so probability of a is equal to one upon two in two seven upon twelve plus one upon two into five upon twelve and This is equal to Let us take one upon two common We have probability of a is equal to one upon two into seven upon twelve plus five upon twelve and This is equal to one upon two into twelve upon twelve and This is equal to one upon two Hence the answer for this question is one upon two That is the probability that the second ball is read is one upon two So this completes our session. I hope the solution is clear to you. Bye and have a nice day