In a game of throwing $3$ coins, a player loses $₹ 5$ for each head and gains $₹ 10$ for each tail. If a random variable $X: S \rightarrow R$ is defined as $X(a) = \text{net gain } (a \in S)$, then the mean of the random variable is (in rupees):

$A$ 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:

If a random variable $X$ has a probability mass function $P(x) = \begin{cases} kx^2, & \text{for } x = 1, 2, 3, 4 \\ 0, & \text{otherwise} \end{cases}$, then the mean of $X$ is ...

If the probability mass function (p.m.f.) of a random variable $X$ is $P(X=x) = \frac{1}{10}$ for $x = 1, 2, 3, \ldots, 10$,and $0$ otherwise,then $\operatorname{Var}(X)$ is equal to:

$A$ coin is tossed three times. Let $X$ denote the number of times a tail follows a head. If $\mu$ and $\sigma^2$ denote the mean and variance of $X$, then the value of $64(\mu+\sigma^2)$ is :

In a city,it is found that $10$ accidents took place in a span of $50$ days. Assuming that the number of accidents follows the Poisson distribution,the probability that there will be $3$ or more accidents in a day in that city is