 Hello and welcome to the session I am Deepika here. Let's discuss a question which says, a die is thrown six times if getting an odd number is a success what is the probability of five successes at least five successes at most five successes. For this question let us first understand Bernoulli trials and binomial distribution. Now trials of a random experiment are called Bernoulli trials if they satisfy the following conditions. One, there should be a finite number of trials. Two, the trials should be independent. Three, each trial has exactly two outcomes success or failure. Four, the probability of success remains the same in each trial. Again, the probability of x successes that is probability of x is equal to x where number of successes is a random variable capital X is also denoted by p x and is given by p x is equal to n c x q raise to power n minus x into p raise to power x where x is equal to 0, 1, 2 so on till n and q is equal to 1 minus p. So, this is a key idea behind that question. We will take the help of this key idea to solve the above question. So, let's start the solution. In this question a die is thrown six times and getting an odd number is a success. So, here the number of trials is finite that is six independent of the outcome of any other trial. So, the trials are independent again each trial has exactly two outcomes success or failure that is getting an odd number or an even number. Now, in this question getting an odd number is a success and getting an even number is a failure and again the probability of success is p is equal to 1 over 2 which is same for all six trials. Hence, according to our key idea the trials of the given experiment Bernoulli trials denote the number of odd numbers experiment of x trials. So, x has a binomial distribution with n is equal to 6, p is equal to 1 over 2. Now, according to our key idea probability of x successes is equal to mcx q raise to power n minus x into p raise to power x where x is equal to 0, 1, 2 so on till n. Now, here is equal to 6, p is equal to 1 over 2 q is equal to 1 minus p and this is equal to 1 minus 1 over 2 which is again equal to 1 over 2. Therefore, probability of x successes is equal to now put here n is equal to 6 and p and q equal to 1 over 2. So, probability of x successes is equal to 6cx 1 over 2 raise to power 6 minus x into 1 over 2 raise to power x and this is again equal to 6cx into 1 over 2 raise to power 6. Now, in part one we have to find the probability of 5 successes. So, probability of 5 successes is equal to 6c5 into 1 over 2 raise to power 6 and this is again equal to 6 factorial over 5 factorial into 6 minus 5 factorial that is 1 factorial into 1 over 2 raise to power 6 and this is again equal to 6 over 2 raise to power 6 which is 64 and this is again equal to 32 and the answer for part one is 3 over 32 that is the probability of 5 successes is 3 over 32. Now, in part two we have to find the probability of at least 5 successes. So, probability of at least 5 successes is given by probability of x greater than equal to 5 and this is equal to probability of 5 successes plus probability of 6 successes. Now, from part one we have probability of 5 successes is 3 over 32. So, this is equal to 3 over 32 plus now probability of 6 successes is given by 6c6 into 1 over 2 raise to power 6. Now, this is equal to 3 over 32 plus now 6c6 is equal to 1. So, 6c6 into 1 over 2 raise to power 6 is 1 over 64. So, this is again equal to 1 over 64 and this is again equal to 7 over 64. Hence, the probability of at least 5 successes is 7 over 64. So, this is the answer for part two. In part three we have to find the probability of at most 5 successes. So, probability of at most 5 successes is given by probability of x less than equal to 5. So, this is equal to probability of 0 successes plus probability of 1 successes plus probability of 2 successes plus probability of 3 successes plus probability of 4 successes plus probability of 5 successes. So, this is equal to 6c0 into 1 over 2 raise to power 6 plus 6c1 into 1 over 2 raise to power 6 plus 6c2 into 1 over 2 raise to power 6 plus 6c3 into 1 over 2 raise to power 6 plus 6c4 into 1 over 2 raise to power 6 plus 6c5 into 1 over 2 raise to power 6. So, this is equal to let us take 1 over 2 raise to power 6 common. So, we have probability of at most 5 successes is equal to 1 over 2 raise to power 6 into 6c0 plus 6c1 plus 6c2 plus 6c3 plus 6c4 plus 6c5. Now, this is again equal to 1 over 64 into now 6c0 is 1 plus 6c1 which is 6 plus 6c2 which is 15 plus 6c3 which is 20 plus 6c4 which is 15 plus 6c5 which is 6. So, this is equal to 63 over 64. Hence the probability of at most 5 successes is 63 over 64. So, this is the answer for part 3. Now, this completes our session. I hope the solution is clear to you. Bye and have a nice day.