 Hello and welcome to the session. The question says three letters are dictated to three persons and an envelope is addressed to each one of them. The letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope. So let's start with the solution. Let three letters be denoted by L1, L2 and L3 and the three envelopes be denoted by E1, E2 and E3. Now the number of ways in which the three letters can be inserted in these three envelopes is three factorial so the number of exhaustive cases is equal to three factorial that is three into two into one which is equal to six. So in six ways these three letters can be inserted in these three envelopes. Suppose we have envelope E1, E2 and E3. Now let us find the possible number of ways in which the letters L1, L2 and L3 can be inserted in these three envelopes. First is L1, L2 and L3. Then we have L1 and L3 have L2, L3, L1 and L1, L3, L2 and L3, L1, L2 and L3, L2 and L1. So the first implies that letter L1 is inserted in envelope E1, letter L2 is inserted in envelope E2 and letter L3 is inserted in envelope E3. Similarly in the second case we have letter L2 is inserted in envelope E1, letter L1 is inserted in envelope E2 and letter L3 in envelope E3. So these are the six possible ways of inserting these three letters in these three envelopes. Let the proper envelope of letter L1 to E1, the proper envelope of letter L2 be E2 and the proper envelope of letter L3 be E3. Now we have to find the probability that at least one letter is in the proper envelope. So first we shall find the number of ways in which one letter is in the correct envelope and two are in the wrong envelope. So the first case is this one, here the letter L3 is in the correct envelope while L2 and L1 are in the wrong envelopes. Then we have this case here the letter L2 is in the correct envelope while L3 and L1 are in the wrong envelopes and then we have L1, here L1 is in the correct envelope E1 while L3 and L2 are in the wrong envelope. So we have three possible ways in which one envelope letter is in the correct envelope while the two other are in the wrong envelopes. So let us write down one letter in correct envelope, two in wrong envelope can be written down as L2, L1, L3 then we have L1, L3, L2 and L3, L2, L1. Similarly on observing we find that all three letters with correct envelope can be stated as L1, L2 and L3. So we have four ways in which at least one letter is in its proper envelope. Therefore probability at least one letter in its proper envelope are four ways that is the number of possible outcomes of four and the total number of possible cases are six which are the exhaustive events and on simplifying we get two upon three. Therefore the probability that at least one letter is in its correct envelope or in other words we can say that more than one letter is in its proper envelope is two upon three. So this completes the session, hope you have understood it Bye and take care