 Hello and welcome to the session let's discuss the following question it says a company knows on the basis of its past experience that 3% of its bulbs are defective. Using Poisson's distribution find the probability that in a sample of 100 bulbs no bulb is defective. So let's now move on to the solution here we have to find the probability that in a sample of 100 bulbs no bulb is defective. So here N is 100 and the probability of success that is P that is the probability of choosing a defective bulb is 3% which is equal to 3 by 100. Now let X denotes a number of defective bulbs. The Poisson's distribution is given by probability of the random variable X at r is given by m to the power r e to the power minus m upon r factorial where m is the mean of the distribution which is given by the formula n into P n is 100 P is 3 by 100. So the mean is 3 now we have to find the probability that no bulb is defective that is we have to find the probability at X is equal to 0 so it is equal to 3 to the power 0 e to the power minus 3 upon 0 factorial 3 to the power 0 is 1 e to the power minus 3 upon 0 factorial is 1 so this is equal to e to the power minus 3 which is given to us as 0.04979 hence the probability that in a sample of 100 bulbs no bulb is defective is 0.04979 so this completes the question and the session you must remember the Poisson's distribution and how do we find mean so this completes the question and the session by for now take care have a good day.