If the probability distribution of a random variable $X$ is given by the following table, then the mean of $X$ is:

$X = x$	$0$	$2$	$4$	$6$	$8$	$10$
$P(X = x)$	$0$	$k$	$2k$	$5k^2$	$2k^2$	$3k$

$A$ service station manager sells gas at an average of ₹ $100$ per hour on a rainy day,₹ $150$ per hour on a dubious day,₹ $250$ per hour on a fair day,and ₹ $300$ per hour on a clear sky day. If weather bureau statistics show the probabilities of weather as follows,then his mathematical expectation is:

Weather	Clear	Fair	Dubious	Rainy
Probability	$0.50$	$0.30$	$0.15$	$0.05$

If the probability density function (p.d.f.) of a continuous random variable $X$ is given by $f(x) = \begin{cases} k(9 + 8x - x^2), & \text{for } -1 \leq x \leq 4 \\ 0, & \text{otherwise} \end{cases}$, then the value of $k$ is:

$A$ random variable $X$ has the following probability distribution:

$X = x_i$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$P(X = x_i)$	$10k$	$9k$	$8k$	$8k$	$6k$	$5k$	$4k$	$3k$	$k$

where $k$ is a real number. If $A = \{ x_i : x_i \text{ is a prime number} \}$ and $B = \{ x_i : x_i > 5 \}$ are two events, then $P(A \cup B) = $

$A$ random variable $X$ has the following probability distribution:

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X=x)$	$k$	$2k$	$3k$	$4k$	$4k$	$3k$	$2k$	$k$	$k$

Then $P(3 < X \leq 6) = $

$A$ random variable $X$ has the following probability density function (p.d.f.):
$f(x) = kx(1-x), 0 \leqslant x \leqslant 1$
If $P(X > a) = \frac{20}{27}$,then find the value of $a$.