 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says The probability that a bulb produced by affecting will fuse after 150 days of use is 0.05 Find the probability that out of five such bulbs, none, not more than one, more than one, at least one will fuse after 150 days of use. Now we know that the trials of a random experiment are Bernoulli trials. If they are finite in number, they are independent. Each trial has exactly two outcomes, success or failure and the probability of success remains the same in each trial. Also, probability of x successes is equal to ncx into q raised to power n-x into p raised to power x where x is from 0 to n and q is equal to 1-p. So this is a key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. Now according to the given question, the probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Since bulbs are taken at random, number of trials is finite, which is fine. The trials are Bernoulli trials. Let x denote the number of bulbs which fuse after 150 days of use. That's the binomial distribution with n equal to 5, which is the probability of success. Now success is when the bulb fuses after 150 days of use and p is given to us 0.05. Therefore q which is given by 1-p is equal to 1-0.05 and this is equal to 0.95. Now according to our key idea, we have probability of x successes is equal to ncx into q raised to power n-x into p raised to power x where x is from 0 till n and q is equal to 1-p. Now here we have n is equal to 5, p is equal to 0.05 and q is equal to 0.95. Therefore probability of x successes is equal to 5cx into 0.95 raised to power 5-x into 0.05 raised to power x. Now in part 1 we have to find the probability that out of 5 bulbs none will fuse after 150 days of use. That is in part 1 we have to find the probability of 0 success. So this is equal to 5c0 into 0.95 raised to power 5-0 into 0.05 raised to power 0 and this is equal to now 5c0 is 1 into 0.95 raised to power 5 into 0.05 raised to power 0 which is 1. So this is equal to 0.95 raised to power 5. Hence the answer for part 1 is 0.95 raised to power 5. Now in part 2 we have to find the probability that out of 5 bulbs not more than 1 will fuse after 150 days of use. So the probability of not more than 1 is given by probability of x is equal to 0 and 1 and this is again equal to probability of 0 success plus probability of 1 success. So this is equal to 5c0 into 0.95 raised to power 5-0 into 0.05 raised to power 0 plus 5c1 into 0.95 raised to power 5-1 into 0.05. 1 into 0.95 raised to power 1 and this is equal to 1 into 0.95 raised to power 5 into 1 less 5c1 which is 5 into 0.95 raised to power 4 into 0.05 raised to power 3 Let us take 0.95 raised to power 4 common from these two terms. So we have probability of not more than 1 is equal to 0.95 raised to power 4 into 0.95 plus 5 into 0.05 which is 0.25 and this is equal to 0.95 raised to power 4 into 1.2. Hence the probability that out of 5 words not more than 1 will fuse after 150 days of use is 0.95 raised to power 4 into 1.2. So this is the answer for part 2. Now in part 3 we have to find the probability that out of 5 words more than 1 will fuse after 150 days of use. So the probability of more than 1 is given by probability of x is equal to 2, 3, 4 and 5. Probability of x is equal to 2, 3, 4 and 5 is equal to 1 minus probability of x is equal to 0 and 1 that is probability of more than 1 is equal to 1 minus probability of not more than 1 and we have the probability that out of 5 words not more than 1 will fuse after 150 days of use is 0.95 raised to power 4 into 1.2. So probability of more than 1 is equal to 1 minus 0.95 raised to power 4 into 1.2. Hence the answer for this part is 1 minus 0.95 raised to power 4 into 1.2. Now in part 4 we have to find the probability that out of 5 words at least 1 will fuse after 150 days of use. So probability of at least 1 bulb fuse is given by probability of x greater than equal to 1 probability of x greater than equal to 1 is given by 1 minus probability of x equal to 0 that is the probability that out of 5 bulbs at least 1 will fuse after 150 days of use is equal to 1 minus probability that out of 5 bulbs none will fuse after 150 days of use and we have probability of x equal to 0 is 0.95 raised to power 5. So this is equal to 1 minus 0.95 raised to power 5. Hence the answer for this part is 1 minus 0.95 raised to power 5. So this completes our session. I hope the solution is clear to you. Bye and have a nice day.