 Hello students, let's solve the following question of probability. It says three points are tossed once. Find the probability of getting three heads, two heads, it leaves two heads, at most two heads, no head, three tails, exactly two tails, no tail, at most two tails. Before moving on to the solution, let us first understand the formula for probability of any event E. If S is the sample space and E is any event, which is the subset of S, then the probability of E is equal to the number of outcomes favorable to E upon the total number of outcomes. So this knowledge will work as the idea. Whereas now proceed on with the solution, we are given that three points are tossed once. So the total number of outcomes will be equal to 2 to the power 3. Because when we toss a coin, there are two possibilities. Either we get a head or we get a tail and we are tossing a coin three times. So it is 2 to the power 3 which is equal to 8. So let us now write the sample space for this experiment. The one of the possibilities that we get head, head, head, that is we get head on all the three coins. Another possibility is that we get head on the first two coins and tail on the last coin. Similarly, we can have HTH or we can have THH, then we can have TTH, then we can have THT. Another possibility is that we get head on the first coin and tail, tail on the last two coins. Then the last possibility is that we get tail on all the coins. So these are eight possibilities. Now in the first part, we have to find the probability of getting three heads. So here event E is getting three heads. From the sample space, we see that there is only one possibility of getting three heads that is HHHH. So the number of outcomes favourable to E is 1. So the probability of getting three heads is equal to the number of outcomes favourable to E upon the total number of outcomes that is 8. So the probability of getting three heads is 1 by 8. Now in the second case, we have to find the probability of getting two heads. So here E is the event of getting two heads. So the number of outcomes favourable to E. Here we can see that 1, 2, 3. There are three outcomes favourable to E, HHHT, HTH, THH is 3, 1, 2, 3. So the probability of getting two heads is equal to the number of outcomes favourable to E upon the total number of outcomes that is 8. So the probability is 3 by 8. Now in the third part, we have to find the probability of getting at least two heads. So here E is the event of getting at least two heads. Now we see that the number of outcomes favourable to E, we want to have at least two heads. That means we can have more than two heads. So one of the possibilities is HHHH. Second one is HHHT. Third one is HHHT. Fourth one is HHHT. And in other cases, we don't get HHHT more than one time. So the number of outcomes favourable to E are 4. So the probability of getting at least two heads is equal to the number of outcomes favourable to E that is 4 upon the total number of outcomes. So the probability is 4 by 8 which is equal to 1 by 2. Now in the fourth part, we have to find the probability of getting at most two heads. So E is the event of getting at most two heads. At most two heads means we should not get more than two heads. We can have one head or we can have two heads. So let's see the sample space and see which are the outcomes favourable to E. These are So the number of outcomes favourable to E are 1, 2, 3, 4, 5 and 6. Since we have to find the probability of getting at most two heads, we can have one more possibility that is we get tail, tail, tail that is we don't get head. So the number of outcomes favourable to E becomes equal to 7, 1, 2, 3, 4, 5, 6 and 7. So the probability of getting at most two heads is equal to the number of outcomes favourable to E that is 7 upon the total number of outcomes. Now we have to find the probability of getting no head. So the number of outcomes favourable to E is only one because this is the only case that is when we get tail, tail, tail on all the coins then we don't get a head. So the probability of getting no head is equal to 1 upon 8. Now in the third, sixth part we have to find the probability of getting three tails. And there is only one possibility of getting three tails that is tail, tail, tail. So the number of outcomes favourable to E is 1. So the probability of getting three tails is equal to 1 upon 8. Now in the seventh part we have to find the probability of getting exactly two tails. So here E is the event of getting at least two exactly two tails. From the sample space we see that which are the possibilities favourable to E we have to get exactly two tails. So this is the one of the possibilities that is tail, tail, head, tail and head, tail, tail. So the number of outcomes favourable to E are three. So the probability of getting exactly two tails is equal to 3 by 8. That is the number of outcomes favourable to E upon the total number of outcomes. Now in the eighth part we have to find the probability of getting no tail. We get no tails if we get all the heads that is the possibility when we get head, head, head. So the number of outcomes favourable to E is 1. So the probability of getting no tail is equal to the number of outcomes favourable to E which comes out to be 1 upon the total number of outcomes. Now in the last part we have to find the probability of getting at most two tails. At most two means we can have zero number of tails or we can have one tail or we can have two tails. But we cannot have more than two tails. So let's see which are the favourable possibilities. So this is the favourable possibility. This one, this one, this one and this we are getting two tails. And this also and this also. So these are seven possibilities. So the number of outcomes favourable to E are seven. So the probability of getting at most two heads its tails is equal to 7 by 8. That is the number of outcomes favourable to E which are seven upon the total number of outcomes. Hence answer to the first part of the question is 1 by 8. To the second part answer is 3 by 8. To the third part answer is 1 by 2. To the fourth part answer is 7 by 8. And to the fifth part answer is 1 by 8. To the sixth part answer is 1 by 8. To the seventh part answer is 3 by 8. To the eighth part answer is 1 by 8. And to the ninth part answer is 7 by 8. And this completes the question and the session. Bye for now. Take care. Have a good day.