 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says There are three coins one is a two-headed coin That is having hacked on both faces Another is a biased coin that comes up heads 75% of the time and third is an unbiased coin One of the three coins is chosen at random and tossed it shows Hems, what is the probability that it was a two-headed coin? Now we will use Bayes theorem to solve this question. So let's start the solution Now according to the question there are three coins one is a two-headed coin Another is a biased coin and the third is an unbiased coin Let even be the event that The coin is two-headed e2 be the event that the coin is biased e3 be the event that the coin is unbiased then probability of e1 is equal to probability of e2 and This is equal to probability of e3 and This is equal to 1 over 3 Again according to the question one of the three coins is chosen at random and tossed it shows heads We have to find the probability that it was a two-headed coin Let a be the event of getting a head probability of a upon e1 that is probability of getting head when First coin is tossed first coin is a two-headed coin that is The coin having head on both faces So probability of a upon e1 is equal to 1 now probability of a upon e2 that is probability of getting head when biased coin is tossed Now the biased coin shows head 75% of the time So probability of a upon e2 is equal to 75 upon 100 Which is equal to 3 over 4 now probability of a upon e3 That is probability of getting head when unbiased coin is tossed is equal to 1 over 2 now We have to find the probability that When one of the three coins is chosen at random and tossed it was a two-headed coin Given it shows head So this is given by probability of even upon a now by using Bayes theorem we have probability of even upon a is equal to probability of even into probability of A upon e1 over probability of even into probability of a upon e1 plus probability of e2 into probability of a upon e2 E2 plus probability of E3 into probability of A upon E3. Now we have probability of E1 is equal to probability of E2 is equal to probability of E3 and this is equal to 1 over 3. Again probability of A upon E1 is 1. Probability of A upon E2 is 3 over 4 and probability of A upon E3 is 1 over 2. So on substituting these values we have probability of E1 upon A is equal to 1 over 3 into 1 upon 1 over 3 into 1 plus 1 over 3 into 3 over 4 plus 1 over 3 into 1 over 2 and this is equal to 1 by 3 upon 1 by 3 plus 1 by 4 plus 1 by 6 and this is equal to 1 by 3 over 4 plus 3 plus 2 over 12 and this is again equal to 1 over 3 into 12 over 9 and this is equal to 4 over 9. So the probability that it was a two-headed coin which shows head is 4 over 9. So this is the answer for the above question. This completes our session. I hope the solution is clear to you. Bye and have a nice day.