In a pizza hut,the following distribution is found for the daily demand of pizzas. Then the expected daily demand and variance are respectively:

No. of Pizzas $(x_i)$	$5$	$6$	$7$	$8$	$9$	$10$
Probability $(P_i)$	$0.07$	$0.2$	$0.3$	$0.3$	$0.07$	$0.06$

The p.m.f of a random variable $X$ is $P(X=x)=\frac{1}{2^5}\binom{5}{x}$,where $x=0, 1, 2, 3, 4, 5$ and $P(X=x)=0$ otherwise. Then:

In a game,a man wins $Rs. 100$ if he gets $5$ or $6$ on a throw of a fair die and loses $Rs. 50$ for getting any other number on the die. If he decides to throw the die either until he gets a five or a six or to a maximum of three throws,then his expected gain/loss (in rupees) is

$A$ boy rolls a die once. If an even number appears, the number of chocolates the boy gets is equal to two more than the number that appeared. If an odd number appears on the die, the number of chocolates he gets is equal to three more than the number that appeared. If a random variable $X$ represents the number of chocolates the boy receives, then the range of $X$ is:

$A$ boy tosses a fair coin $3$ times. If he gets $₹ 2x$ for $x$ heads,then his expected gain equals to $₹........$

If a Poisson variate $X$ satisfies $P(X=2) = P(X=3)$, then $P(X=5) =$