 Good morning friends and poor one today. We will discuss the following question from a lot of 30 bulbs Which includes six defectives a sample of four bulbs is drawn at random with replacement? Find the probability distribution of the number of defective bulbs Let us begin with the solution now Now let X denotes the number of defective bulbs drawn in four samples Then X can take values 0 1 2 3 4 as there can be 0 1 2 3 4 Bulbs drawn which are defective That is the maximum number of defective bulbs that can be drawn as 4 and the minimum number is 0 now probability of drawing a defective bulb is 6 upon 30 as there are six defective bulbs and in all there are 30 bulbs This is equal to 1 upon 5 So the probability of drawing non defective bulbs is 1 minus 1 upon 5 and we get this is equal to 4 upon 5 At X is equal to 0 We have probability of drawing zero defective bulbs that is probability of drawing four non defective bulbs in four draws is 4 upon 5 Into 4 upon 5 into 4 upon 5 Into 4 upon 5 because probability of drawing non defective bulbs is 4 upon 5 And we have this is equal to 256 upon 625 At X is equal to 1 We have probability of drawing one defective bulb is equal to probability that first bulb is defective that is 1 upon 5 into probability that we have three non defective bulbs and it is given by 4 upon 5 into 4 upon 5 into 4 upon 5 plus Probability of drawing one non defective bulb that is 4 upon 5 Into probability of drawing a defective bulb that is 1 upon 5 Into probability of drawing two non defective bulbs, which is 4 upon 5 into 4 upon 5 plus Probability of drawing two non defective bulbs that is 4 upon 5 into 4 upon 5 Into probability of drawing a defective bulb that is 1 upon 5 Into probability of drawing one non defective bulb that is 4 upon 5 plus Probability of drawing three non defective bulbs that is 4 upon 5 into 4 upon 5 into 4 upon 5 into probability of drawing one defective bulb that is one upon five. So probability of drawing one defective bulb has four arrangements. That is the first bulb is defective and others are non-defective or the second bulb is defective whereas others are non-defective or third bulb is defective whereas the other three are non-defective or the fourth bulb is defective whereas the other three are non-defective. So we can write this as this is equal to 4 into 4 upon 5 into 4 upon 5 into 4 upon 5 into 1 upon 5 and we get this is equal to 256 upon 625. Now at x is equal to 2 the probability of drawing two defective bulbs is 6 into probability of drawing one non defective bulb that is 4 upon 5 into probability of drawing one non defective bulb that is 4 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that is 1 upon 5 into probability of drawing a defective bulb that upon 5 as there can be 6 arrangements for drawing 2 defective bulbs so we get 6 into 4 upon 5 into 4 upon 5 into 1 upon 5 into 1 upon 5 and we get this is equal to 96 upon 625. Now at x is equal to 3 we have probability of drawing 3 defective bulbs is equal to 4 into probability of drawing 3 defective bulbs that is 1 upon 5 into 1 upon 5 into 1 upon 5 into probability of drawing 1 non defective bulb that is 4 upon 5. As probability of drawing 3 defective bulbs has 4 arrangements so we get 4 into 1 upon 5 into 1 upon 5 into 1 upon 5 that is probability of drawing 3 defective bulbs into probability of drawing 1 non defective bulb that is 4 upon 5 and we get this is equal to 16 upon 625. Now finally at x is equal to 4 we have probability of drawing 4 defective bulbs is equal to 1 upon 5 into 1 upon 5 into 1 upon 5 into 1 upon 5 as we have probability of drawing a defective bulb is 1 upon 5 and we get this is equal to 1 upon 625. So now we will draw the probability distribution table. So this is our probability distribution table. Here we have when x is equal to 0 1 2 3 and 4 then we have probability of x as 256 upon 625 256 upon 625 96 upon 625 16 upon 625 and 1 upon 625 respectively. This is our answer. Hope you have understood the solution. Bye and take care.