 Hello and welcome to the session. In this session we discuss the following question which says assume that when a salesperson makes phone calls 1 out of 11 will be busy. What is the probability that after 8 numbers are called at least 3 of them will be busy. The probability of the random variable x equal to r is equal to ncr into p to the power r into q to the power n minus r where we have this r is equal to 0, 1, 2 and so on up to n where n is the number of trials r are the number of successes then n minus r would be the number of failures p is the probability of getting success q is the probability of getting failure. This is the key idea that we use for this question. Let's proceed with the solution now. We take let p be the probability that the number is busy. Now since in the question it's given that when a salesperson makes phone calls 1 out of 11 will be busy. So we say that p that is the probability that the number is busy is equal to 1 upon 11. We know that p plus q is equal to 1. So this means q is equal to 1 minus p that is q is equal to 1 minus 1 upon 11 which gives us q equal to 10 upon 11. n is the number of trials and in this case n would be equal to 8 since in the question it's given that 8 numbers are called. We take let x be the random variable the number called is busy and we have to find the probability that after 8 numbers are called at least 3 of them will be busy. So that means we need to find the probability when the random variable x is greater than equal to 3. Now probability when the random variable x is greater than equal to 3 is equal to 1 minus the probability when the random variable x is less than 3. So this is equal to 1 minus probability when the random variable x is equal to 0 plus the probability when the random variable x is equal to 1 plus the probability when the random variable x is equal to 2. Using the formula given in the key idea to find the probability of the random variable x equal to r that is this formula we find the probability of x equal to 0, x equal to 1, x equal to 2. So we can write probability when x greater than equal to 3 is equal to 1 minus now the probability of x equal to 0 is written as 8 c 0 that is n is 8. So in place of n we write 8 into p which is equal to 1 upon 11 to the power r which is 0 in this case into q which is 10 upon 11 to the power n minus r that is 8 minus 0 which is 8. In the same way we can write the probability of x equal to 1 and probability of x equal to 2. So this would be plus 8 c 1 into 1 upon 11 whole to the power 1 into 10 upon 11 whole to the power 8 minus 1 that is 7 plus 8 c 2 into 1 upon 11 whole to the power 2 into 10 upon 11 whole to the power 8 minus 2 that is 6. So this is further equal to 1 minus now 8 c 0 would be 1 into 1 into 10 upon 11 whole to the power 8 plus 8 into 1 upon 11 into 10 upon 11 whole to the power 7 plus 28 into 1 upon 11 whole to the power 2 into 10 upon 11 whole to the power 6 and so this is equal to 1 minus 10 upon 11 whole to the power 8 plus 8 upon 11 into 10 upon 11 whole to the power 7 plus 28 upon 121 into 10 upon 11 whole to the power 6. So we get this is further equal to 1 minus 10 upon 11 whole to the power 6 common into 10 upon 11 whole to the power 2 plus 8 upon 11 into 10 upon 11 plus 28 upon 121. This further is equal to 1 minus 10 upon 11 whole to the power 6 into 100 upon 121 plus 80 upon 121 plus 28 upon 121 which further is equal to 1 minus 10 upon 11 whole to the power 6 this whole into 100 plus 80 plus 28 upon 121 which gives us 1 minus 10 upon 11 whole to the power 6 and this whole into 208 upon 121. Thus we get probability when the random variable x is greater than equal to 3 is equal to 1 minus 10 upon 11 whole to the power 6 into 208 upon 121. So this is the required probability this is our final answer. So this completes the session. Hope you have understood the solution of this question.