 Hi friends, I am Purvan. Today we will discuss the following question. A fair coin and an unbiased die are tossed. Let A be the event head appears on the coin and B be the event three on the die. Check whether A and B are independent events or not. Let E and F be two events. Then E and F are independent. If probability of E intersection F is equal to probability of E into probability of F. So this is the key idea behind our question. Let us begin with the solution now. Now let S be the sample space of the experiment of tossing a fair coin and an unbiased die. Now when a coin is tossed we can get only two outcomes either a head or a tail. And when a die is thrown we can get six outcomes that is either one, two, three, four, five or six appear on the top face. So when a fair coin and an unbiased die are tossed we can get two into six that is twelve outcomes. And those twelve outcomes will be head and one, head and two, head and three, head and four, head and five, head and six, tail and one, tail and two, tail and three, tail and four, tail and five, tail and six. So these are the twelve outcomes which we get when a fair coin and an unbiased die are tossed. Now we are given that A is the event that head appears on the coin and B is the event that three appears on the die. Then probability of A that is probability of head appearing on the coin will be six upon twelve. Because out of the twelve outcomes there are six outcomes in which we get head when a fair coin and an unbiased die are tossed and this is equal to one upon two. And probability of B that is probability that three appears on the die is equal to two upon twelve. Now out of these twelve outcomes there are only two outcomes in which three appears on the die. One is when head appears on the coin and one is when tail appears on the coin. So we get probability of B is equal to two upon twelve which is equal to one upon six. Now A intersection B is equal to a set which consists of elements which appear both in A and in B. Now A is the event that head appears on the coin and B is the event that three appears on the die. Now out of these outcomes there is only one outcome in which head appears on the coin and three appears on the die. So we get A intersection B is equal to a set which consists of only one outcome that is head and three. So we get probability of A intersection B is equal to one upon twelve. Also probability of A into probability of B is equal to probability of A is equal to one upon two into probability of B is equal to one upon six. And we get this is equal to one upon twelve. Now we mark this as one and we mark this as two. So from 1 and 2 we have probability of A intersection B is equal to probability of A into probability of B. Now by key idea we know that two events E and F are independent if probability of E intersection F is equal to probability of E into probability of F. Thus from key idea we conclude that the events A and B are independent. Thus we write our answer as A and B are independent. Hope you have understood the solution. Bye and take care.