 Hello and welcome to the session. In this session, we discussed the following question which says, some a lot of 10 bulbs, which includes three defectives. A sample of two bulbs is drawn at random. Find the probability distribution of the number of defective bulbs. Before we move on to the solution, let's discuss the probability distribution in variable given by x and the probability distribution is given as, takes the values as x1, x2, up to xn. When p of x, that is the probability of the random variable x, takes the values on up to pn. The probability distribution of the random variable x is m0 and the solution of from 1 to n is equal to 1 and the real numbers, possible values, random variable, probability, taking the value xi. So xi is equal to pi. So this is how we give the probability distribution of a random variable. This is the key idea that we use for this question. Let's move on to the solution. As with the question we have, that we are given a lot of 10 bulbs out of which three bulbs are defective. So this means 10 minus 3, 7 bulbs would be good bulbs. And from this lot of 10 bulbs, a sample of two bulbs is drawn at random. We have to find the probability distribution of the number of defective bulbs. So we take let x be the number of defective bulbs in the sample space. In this case, our x is the random variable. We have to find the probability distribution of this x, which is the number of defective bulbs in the sample space. As given in the key idea, we have that these real numbers x1, x2, up to xn are the possible values of the random variable x. So let's see what would be the possible values of the random variable x. Now as we draw two bulbs from the lot of 10 bulbs, so these two bulbs could be defective or either of the two bulbs would be defective or minus 10 would be defective. So this means the random variable values 0, 1 or 2, therefore we have so x1 equal to 0, x2 equal to 1 and x3 equal to 2. These are the three possible values of the random variable x. Now for the probability distribution, we have got the possible values of the random variable x. Now we have left two probability of the random variable x at each of these values of x1, x2 and x3. The total number of bulbs are given as would be good bulbs. To consider x1 equal to 0, let's find out the probability of the random variable x equal to x1 that is 0. This would be the probability that one of the drawn bulbs are defective. From which we draw, I'll give good bulbs 3 equal to c2. So there are seven good bulbs and out of the seven good bulbs we draw two bulbs. So 7 c2 equal to total of 10 bulbs out of which we draw two bulbs. This would be equal to 7 c2 which is 5 factorial upon into 10 minus 2 that is 8 factorial. Further we get this is equal to upon 2 into 1, the small upon and 3, 3 times is 9 and 2, 5 times is 10. So this is equal to the probability of the random variable a equal to 7 upon 15. The probability of the random variable, so probability of x equal to 1 would be the same as the probability of defective bulbs. So 3 c1, since we draw one defective bulb out of the three defective bulbs and this into and we draw one good bulb and this one upon 10 c2 and we draw two bulbs from the lot of 10 bulbs. Equal to 3 factorial upon 1 factorial into 3 minus 3, 3 times is 9 into 3 that is 15. So probability of the random variable 2. So we now find out the probability of the random variable x equal to x3 which is 2. This would be same as the probability of v equal to 3 factorial into 3 minus 2 that is 1 factorial this whole upon 10 factorial 6 factorial. This gives us 3 this one upon 10 into 9 upon this is 9. So this is equal to 15 minus we have probability of the random variable x equal to 15. The probability distribution p values for the random variable x as 0 1 to 0 the probability is given as 7 upon 15. We write here 7 upon 15 the probability is given as 7 upon 15. So again we write here 15 to do 2 the probability is given as 1 upon 15. So here we have 1 upon this we require probability distribution of the random variable x where this x is the number of defective bulbs in the sample space. So this concludes the session. Hope you have understood the solution of this question.