 Hi, and welcome to the session. Let's discuss the following question. The question says from a lot of 30 bulbs which include 6 defective bulbs, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs. Let's now start the solution. In this question, we have to find the probability distribution of the number of defective bulbs. So, let's denote the number defective bulbs. Bulbs are drawn at random. So, this implies that x can clearly take the value 0, 1, 2, 3 and 4. Now, we will find the probability of x at 0, 1, 2, 3 and 4. We will now find the probability of getting a defective bulb. Number of bulbs is 30 out of 6 are defective. So, probability of getting defective bulb is 6 by 30 and this is equal to 1 by 5. Now, we will find the probability of getting a non-defective bulb. Number of non-defective bulbs is 24 and total number of bulbs is 30. So, probability of getting a non-defective bulb is 24 by 30 and this is equal to 4 by 5. Now, we will find the probability of x at 0, 1, 2, 3 and 4. So, let's first find probability of x at 0. Probability of x equals to 0 means we have to find the probability of getting non-defective bulb. That means in all the four draws, bulbs are not defective and we know that probability of getting a non-defective bulb is 4 by 5. So, probability of x equals to 0 is 4 by 5 into 4 by 5 into 4 by 5 into 4 by 5 and this is equal to 256 upon 625. Now, we will find the probability of x at 1. Probability of x equals to 1 means we have to find the probability of getting 1 defective bulb and 3 non-defective bulb. Now, this 1 defective bulb can be in the first draw, second draw, third draw or in the fourth draw. So, when the first bulb drawn is defective and other 3 bulbs are non-defective then the probability is 1 by 5 into 4 by 5 into 4 by 5 into 4 by 5. But when the second bulb drawn is defective then the probability is 4 by 5 into 1 by 5 into 4 by 5 into 4 by 5 and when the third bulb drawn is defective then the probability is 4 by 5 into 4 by 5 into 1 by 5 into 4 by 5 and when the last bulb drawn is defective then the probability is 4 by 5 into 4 by 5 into 4 by 5 into 1 by 5 and this is equal to 4 times 1 by 5 into 4 by 5 into 4 by 5 into 4 by 5 and this is equal to 256 upon 625. So, probability of x at 1 is 256 upon 625. Now, we will find the probability of x at 2. Now, probability of x equals to 2 means we have to find the probability of getting 2 defective bulbs and 2 non-defective bulbs. When the first two bulbs are defective and other two are non-defective then the probability is 1 by 5 into 1 by 5 into 4 by 5 into 4 by 5 plus when the second and third bulb is defective and first and fourth are non-defective then the probability is 4 by 5 into 1 by 5 into 1 by 5 into 4 by 5 plus when the last two bulbs are defective then the probability is 4 by 5 into 4 by 5 into 1 by 5 into 1 by 5 plus when the first and the last bulb is defective then the probability is 1 by 5 into 4 by 5 into 4 by 5 into 1 by 5 plus when the second and the last bulb is defective then the probability is 4 by 5 into 1 by 5 into 4 by 5 into 1 by 5 plus when the first and the third bulb is defective then the probability is 1 by 5 into 4 by 5 into 1 by 5 into 4 by 5 and this is equal to 6 times 1 by 5 into 1 by 5 into 4 by 5 into 4 by 5 and this is equal to 96 upon 625 we will find the probability of x at 3 now x equals to 3 means we have to find the probability of getting 3 defective bulbs and 1 non-defective bulb when first three bulbs are defective and last one is non-defective then the probability is 1 by 5 into 1 by 5 into 1 by 5 into 4 by 5 when the last three bulbs are defective and the first one is non-defective then the probability is 4 by 5 into 1 by 5 into 1 by 5 into 1 by 5 plus when the first third and fourth bulb are defective then the probability is 1 by 5 into 4 by 5 into 1 by 5 into 1 by 5 plus when the first second and the last bulb is defective then the probability is 1 by 5 into 1 by 5 into 4 by 5 into 1 by 5 and this is equal to this is equal to 4 times 1 by 5 into 1 by 5 into 1 by 5 into 4 by 5 this is equal to 16 upon 625 and now finally we will find the probability of x at 4 now x equals to 4 means we have to find the probability of getting all defective bulbs that means in all the four draws bulbs are defective so the probability is 1 by 5 into 1 by 5 into 1 by 5 into 1 by 5 and this is equal to 1 upon 625 now we will write the probability distribution probability distribution is the tabular representation of the values of a random variable x and its corresponding probability when x is equal to 0 then its probability is 256 upon 625 when x is equal to 1 then its probability is 256 upon 625 when x is equal to 2 then its probability is 96 upon 625 when x is equal to 3 then its probability is 16 upon 625 and when x is equal to 4 then its probability is 1 upon 625 this is our required answer bye and take care hope you have enjoyed the session