 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says A die is tossed thrice. Find the probability of getting an odd number at least once. So let's start the solution. Now according to the question, a die is tossed thrice. So the sample space of the given experiment has 216 elements. Now we have to find the probability of getting an odd number at least once. Let E be the event of getting an odd number at least once. So E complement will be the event of getting no odd numbers. Now we will first write down the elements of the set E complement. So the elements of the set E complement are of the form 222 that is 2 in all the 3 tosses. Again 2 in the first 2 tosses and 4 in the second toss. Similarly 2 in the first 2 tosses and 6 in the third toss. Again suppose the die shows number 2 in the first toss and number 4 in the second toss and number 2, 4 and 6 in the third toss. So we have 242, 244 and 246. Similarly suppose the die shows number 2 in the first toss, number 6 in the second toss and numbers 2, 4 and 6 in the third toss. So we have 262, 264 and 266. Now these are the outcomes when the die shows number 2 in the first toss. So these are 9 outcomes. Similarly we have 9 outcomes when the die shows number 4 in the first toss. Also we have 9 more outcomes when the die shows number 6 in the first toss. So the set E complement has 27 elements. Therefore the number of outcomes favorable to event E complement is equal to 27. Now we know that probability of E that is getting an odd number at least once is equal to 1 minus probability of getting no odd number that is probability of E complement and this is equal to 1 minus number of outcomes favorable to the event E complement that is 27 upon total number of outcomes which is 216 and this is equal to 1 minus 1 over 8 and this is equal to 7 upon 8. Hence the probability of getting an odd number at least once is equal to 7 upon 8. So the answer for this question is 7 upon 8. I hope the solution is clear to you. Bye and have a nice day.