In a meeting,$70 \%$ of the members favour and $30 \%$ oppose a certain proposal. $A$ member is selected at random. We take $X=0$ if he opposes the proposal and $X=1$ if the member is in favour. Then the variance of $X$ is:

If $X$ is a Poisson variate such that $2 P(X=1)=5 P(X=5)+2 P(X=3)$,then the standard deviation of $X$ is

State which of the following is not the probability distribution of a random variable. Give reasons for your answer.

$Y$	$-1$	$0$	$1$
$P(Y)$	$0.6$	$0.1$	$0.2$

Let the mean and the standard deviation of the probability distribution be $\mu$ and $\sigma$,respectively. If $\sigma - \mu = 2$,then $\sigma + \mu$ is equal to:

$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

If $X$ is a Poisson variable representing the number of successes in $50$ trials such that $2 P(X=1) = 5 P(X=5) + 2 P(X=3)$, then the probability of getting success in one trial is

$A$ person plays a game of tossing a coin thrice. For each head,he is given $Rs. 2$ by the organiser of the game and for each tail,he has to give $Rs. 1.50$ to the organiser. Let $X$ denote the amount gained or lost by the person. Show that $X$ is a random variable and exhibit it as a function on the sample space of the experiment.