The discrete random variables $$X$$ and $$Y$$ are independent from one another and are defined as $$X \sim B(16,0.25)$$ and $$Y \sim P(2)$$. Then, the sum of the variance of $$X$$ and $$Y$$ is

If 6 is the mean of a Poisson distribution, then $$P(X \geq 3)=$$

A coin is tossed until a head appears or it has been tossed thrice. Given that head doesn’t appear on the first toss, the probability that coin tossed thrice is

Box-I contains 3 cards bearing numbers 1, 2, 3 , Box II contains 5 cards bearing numbers 1 , 2, 3, 4, 5 and Box III contains 7 cards bearing numbers 1, 2, 3, 4, 5, 6, 7. One card is drawn at random from each of the boxes. If $$x_i$$ be the number on the card drawn from the $$i$$ th box, $$i=1,2,3$$, then the probability that $$x_1+x_2+x_3$$ is odd is equal to