 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed? Let us now start with the solution. First of all, let us assume that variable x denotes the number of right-handed persons in a random sample of 10 people. Now we know out of 10 people, each person can be right-handed or left-handed. And we also know that trials made for selection of right-handed persons in a random sample of 10 people are independent of each other. So these are Bernoulli trials. So we can write repeated selection of people are Bernoulli trials. Now clearly we can see variable x has binomial distribution with n is equal to 10 and p is equal to 90 upon 100. That is 0.9. We know total number of persons selected is equal to 10 and we also know that 90% of the people are right-handed. So probability that a person selected is a right-handed person is equal to 90 upon 100. That is 0.9. Therefore probability of x successes is equal to ncx q raised to the power n-x p raised to the power x where p is equal to 0, 2. Now here in the given question we have n is equal to 10, p is equal to 0.9, q is equal to 1-p that is 1-0.9 which is further equal to 0.1. Here q represents the failure that is the person selected is not a right-handed person. Now substituting corresponding values of n, p and q in this expression we get probability of x successes is equal to 10cx 0.1. raised to the power 10-x 0.9 raised to the power x. Clearly we can see here we have replaced n by 10, q by 0.1 and p by 0.9. Now we have to find the probability that at most 6 persons right-handed. Or we can say we have to find the probability of x less than equal to 6. Now we know probability of x less than equal to 6 is equal to 1-probability of x greater than 6. Now this is further equal to 1-probability of x is equal to 7 plus probability of x is equal to 8 plus probability of x is equal to 9 plus probability of x is equal to 10. Now we can find value of probability of x is equal to 7 by substituting 7 for x in right-hand side of this expression and we get 1-10c7 0.1 raised to the power 3.9 raised to the power 7. Similarly we can find the value of probability of x is equal to 8 by substituting 8 for x in right-hand side of this expression and we get 10c8 0.1 raised to the power 2 multiplied by 0.9 raised to the power 8. Now we will find the value of probability of x is equal to 9. We will substitute 9 for x in this expression and we get 10c9 0.1 raised to the power 10-9 that is 1 multiplied by 0.9 raised to the power 9 plus Now we will find the probability of x is equal to 10 by substituting 10 for x in right-hand side of this expression and we get 0.9 raised to the power 10. We know 10c10 is equal to 1, 10-10 is equal to 0 and anything raised to the power 0 is equal to 1 only and we are left with 0.9 raised to the power 10. Now this expression can be further written as 1 minus summation of 10cr 0.9 raised to the power r multiplied by 0.1 raised to the power 10-r where r is equal to 7 to 10. We know we can represent the sum of these four terms by this expression. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.