 Hello and welcome to the session. In this session we discussed the following question that says the probability function of a discrete random variable capital X is given by probability of capital X equal to x is equal to lambda upon 2 to the power of x where x takes the values from 1, 2, 3 and so on. Find first part the value of lambda, second part probability of capital X less than equal to 7. If a random variable capital X takes the values x1, x2, x3 and so on and so on up to xn with probabilities px1, px2, px3 and so on up to pxn where we have summation of pxi where i goes from 1 to n is equal to 1 where this p is the probability density function of the random variable X. This is the key idea that we use in this question. Let's now move on to the solution. In the question we have a probability function of a discrete random variable X defined as probability of the random variable X equal to x is equal to lambda upon 2 to the power of x where x takes the values from 1, 2, 3 and so on and in the first part we have to find the value of lambda. So first we have the probability of the random variable X equal to x is equal to lambda upon 2 to the power of x where x takes the values from 1, 2, 3 and so on and in the first part here we need to find the value of lambda. Now since we know that summation of the probabilities is equal to 1, so summation of the probability of x where x goes from 1 to infinity is equal to 1. Now this means summation lambda upon 2 to the power of x where x goes from 1 to infinity is equal to 1. Now putting the values of x as 1, 2, 3 and so on up to infinity in this and doing its summation equal to 1 we have lambda upon 2 plus lambda upon 2 square plus lambda upon 2 cube plus and so on plus lambda upon 2 to the power infinity is equal to 1. Now further taking lambda common we have lambda into 1 upon 2 plus 1 upon 2 square plus 1 upon 2 cube plus and so on plus 1 upon 2 to the power of infinity is equal to 1. This is an infinite GP and we know that sum of an infinite GP is equal to a upon 1 minus r which in this case would be equal to a that is the first term which is 1 upon 2 upon 1 minus the common ratio which is again 1 upon 2 this is equal to 1 upon 2 upon 1 upon 2 equal to 1. So here we have lambda into 1 is equal to 1 this gives us the value of lambda as 1. This solves our first part now in the second part we have to find the probability of the random variable x less than equal to 7. Now the probability of the random variable x less than equal to 7 would be equal to the probability of the random variable x equal to 1 plus the probability of the random variable x equal to 2 plus the probability of of the random variable x equal to 3 plus the probability of the random variable x equal to 4 plus probability of random variable x equal to 5 plus the probability of the random variable x equal to 6 plus the probability of the random variable x equal to 7. Now this probability would be same as this that is lambda upon 2 to the power of x, this gives us lambda upon 2 to the power of 1 that is we put the value of x as 1 plus lambda upon 2 to the power of 2 plus lambda upon 2 to the power of 3 plus lambda upon 2 to the power of 4 plus lambda upon 2 to the power of 5 plus lambda upon 2 to the power of 6 plus lambda upon 2 to the power of 7 this is the probability of the random variable x less than equal to 7 now as we have lambda equal to 1 so we can say that the probability of the random variable x less than equal to 7 is equal to 1 upon 2 plus 1 upon 2 square plus 1 upon 2 cube plus 1 upon 2 to the power 4 plus 1 upon 2 to the power 5 plus 1 upon 2 to the power 6 plus 1 upon 2 to the power 7 now we know that some of the n terms of a GP is equal to a into 1 minus r to the power of n upon 1 minus r now this is a GP of 7 terms and so its sum would be equal to a that is 1 upon 2 which is the first term into 1 minus r which is the common ratio and it is 1 upon 2 whole to the power of n which is 7 in this case and this whole upon 1 minus the common ratio which is 1 upon 2 so the probability of the random variable x less than equal to 7 is equal to 1 upon 2 into 1 minus 1 upon 2 to the power of 7 this whole upon 1 upon 2 now this 1 upon 2 1 upon 2 cancels and so this is equal to 2 to the power of 7 minus 1 upon 2 to the power of 7 further this is equal to 128 minus 1 upon 128 which is equal to 127 upon 128 so this is the probability of the random variable x less than equal to 7 this is the answer for the second part of the question this completes the session hope you understood the solution of this question