 Hello and welcome to the session. In this session we will discuss the following question which says a random variable x has the following probability distribution in which we are given the values of x as 0, 1, 2, 3, 4, 5, 6 and the corresponding probabilities of the random variable x are 0, 2k, k, 6k, 5k, k, 4k respectively. In the first part we have find the value of k in the second part 5 probability of the random variable x less than 5 probability of the random variable x greater than equal to 5 and probability of the random variable x greater than 0 and less than 4. In the third part we have give the smallest value of m for which probability of the random variable x less than equal to m is greater than equal to 0.7. Consider a random variable capital x which takes the values say x1, x2 and so on up to xn and this random variable x has probabilities at x1, x2 and so on up to xn as px1, px2 and so on up to pxn where the summation pxi where i goes from 1 to n is equal to 1. Each of these probabilities that is probability of xi is greater than 0. Then this p is called the probability density function of the random variable x. This is the key idea that we use in this question. Let's proceed with the solution now. Now in the question we are given the probability distribution of the random variable x as this and in the first part we are supposed to find the value of k. For different values of x we are given the probabilities also. Now as we know that some of these probabilities is equal to 1. So in the first part we have summation of px that is probability of the random variable x where x takes the values from 0 to 6. This would be equal to 1. So now we will add these values of the probabilities for different values of x from 0 to 6. So this means we have 0 plus 2k plus k plus 6k plus 5k plus k plus 4k is equal to 1. This gives us 19k is equal to 1 and from here we have the value of k as 1 upon 19. So this is the answer for the first part of the question. Now let's move on to the second part. In the second part we are supposed to find the probability of the random variable x less than 5. So first of all let's see how we find this. Now probability of random variable x less than 5 is equal to probability of the random variable x equal to 0 plus probability of the random variable x equal to 1 plus probability of the random variable x equal to 2 plus probability of the random variable x equal to 3 plus the probability of random variable x equal to 4. Now the probability of the random variable x at 0 is 0. So here we have 0 plus. Now let's see what is the probability of the random variable x equal to 1. So when x is equal to 1 the probability of the random variable is 2k. So next we add 2k then we have the probability of the random variable x equal to 2. When x is equal to 2 its probability is k. So here we add k plus now the probability of the random variable x equal to 3. Now when x is equal to 3 its probability is 6k. So we add 6k here plus the probability of the random variable x equal to 4. Now when x is equal to 4 its probability is 5k. So here we have 5k. Now this would be equal to 14k. So now we know that k is equal to 1 upon 19. Now putting the value of k as 1 upon 19 we get 14 upon 19 as the probability of the random variable x less than 5. Now let's see what we are supposed to find next. Next we have to find the probability of the random variable x greater than equal to 5. Now let this be a next we will find the v part that is probability of the random variable x greater than equal to 5. This would be equal to 1 minus probability of the random variable x less than 5. So this is equal to 1 minus 14 upon 19 which is equal to 19 minus 14 that is 5 upon 19 this is the probability of the random variable x greater than equal to 5. Next we will find the probability of the random variable x greater than 0 and less than 4. So now the c part of this question is probability of the random variable x greater than 0 less than 4. This is equal to probability of the random variable x equal to 1 plus probability of the random variable x equal to 2 plus probability of the random variable x equal to 3. Now from the table of the probability distribution of the random variable x we have the probability of the random variable x at 1 is 2k at 2 is k and at 3 is 6k. So this would be equal to 2k which is the probability of the random variable x equal to 1 plus k which is the probability of the random variable x equal to 2 plus 6k which is the probability of the random variable x equal to 3 and this is equal to 9k. Now the value of k is 1 upon 19 so here we have 9 upon 19 this is the probability of the random variable x greater than 0 and less than 4. Now in the third part of the question we have to give the smallest value of m for which we have the probability of the random variable x less than equal to m would be greater than equal to 0.7. So now we have the third part. Now we need to find the smallest value that probability of the random variable x less than equal to m is greater than equal to 0.7. Now as we have already found out the probability of the random variable x greater than 0 and less than 4 is 9 upon 19. So from here we can say that the probability of the random variable x less than 4 is equal to 9 upon 19 and this is equal to 0.47. So probability of the random variable x less than 5 which is equal to as already found out 14 upon 19 is equal to 0.74. Now we can also say that this is equal to probability of the random variable x less than equal to 4 that is probability of the random variable x less than 5 is same as the probability of the random variable x less than equal to 4 and this is equal to 0.74. So from here we can say that the smallest value of m is 4. Probability of the random variable x less than equal to 4 is equal to 0.74 and it satisfies this condition that is probability of the random variable x less than equal to m is greater than equal to 0.7. So here we have the value of m as 4. So this completes the session. Hope you have understood the solution of this question.