1.Let XXX be a random variable with the probability distribution P(X=0)=1mP(X = 0) = \dfrac{1}{m}P(X=0)=m1, P(X=j)=2m⋅2jP(X = j) = \dfrac{2}{m \cdot 2^{j}}P(X=j)=m⋅2j2 for j=1,2,3,…,∞j = 1, 2, 3, \ldots, \inftyj=1,2,3,…,∞. Then the mean of the distribution and P(X is positive and even)P(X \text{ is positive and even})P(X is positive and even), respectively, are:a.43\dfrac{4}{3}34 and 19\dfrac{1}{9}91b.43\dfrac{4}{3}34 and 29\dfrac{2}{9}92c.34\dfrac{3}{4}43 and 19\dfrac{1}{9}91d.34\dfrac{3}{4}43 and 29\dfrac{2}{9}92Login to continueOnly logged in users canattempt or see the solution.
