Find the mean number of heads in three tosses of a fair coin.

If a random variable $X$ follows the Poisson distribution with variance $3$, then $P(X=r)$ is maximum, when $r=$

$A$ random variable $X$ has the following probability distribution:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is:

$A$ player tosses two coins. He wins $Rs. 1$ if $1$ head appears,$Rs. 2$ if $2$ heads appear. But he loses $Rs. 3$ if no head appears. The mean of the prize money is

The random variable $X$ has the following probability distribution:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ is:

If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is:

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$