 Hello and welcome to the session I am Deepika here. Let's discuss a question which says determine probability of E upon F when a coin is tossed three times where E is the event at most two tails and F is the event at least one tail. So let's start the solution. Let S be the sample space of the experiment of tossing a coin three times. So the elements of SR that is S is equal to HTT that is hat on the first toss and tails on the second and third toss. Similarly H H T H H H that is all the three tosses results into hat H T H T T H T T that is tails on all the three tosses T H T and T H H. Now according to the question when a coin is tossed three times E is the event there is at most two tails and F is the event there is at least one tail. So elements of ER H T T H H T H H H T H and T H T and T H T and F is equal to H T T H H T H T H H T H T T T H H. All the elements of E and F are common except these two elements. So E intersection F is equal to H T T H H H T H T H and T H H probability of E is equal to 7 upon 8 because E has seven elements and the sample space has eight elements and probability of F is also 7 upon 8. Again the probability of E intersection F is equal to 6 upon 8. Now we know that probability of E upon F is equal to probability of E intersection F upon probability of F provided probability of F is not equal to zero. So this is equal to 6 upon 8 into 8 upon 7 and this is again equal to 6 upon 7. Hence the answer for this question is 6 upon 7. So this completes our session. I hope the solution is clear to you. Bye and have a nice day.