In a game,$3$ coins are tossed. $A$ person is paid ₹ $100$,if he gets all heads or all tails; and he is supposed to pay ₹ $40$,if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) is

If a continuous random variable $X$ has probability density function $f(x)$ given by $f(x) = \begin{cases} ax, & 0 \le x < 1 \\ a, & 1 \le x < 2 \\ 3a - ax, & 2 \le x \le 3 \\ 0, & \text{otherwise} \end{cases}$,then $a$ has the value:

At a telephone enquiry system,the number of phone calls regarding relevant enquiries follows a Poisson distribution with an average of $5$ phone calls during $10$-minute time intervals. The probability that there is at most one phone call during a $10$-minute time period is:

$A$ random variable $X$ takes values $0, 1, 2, 3$ with probabilities $\frac{2a + 1}{30}, \frac{8a - 1}{30}, \frac{4a + 1}{30}, b$ respectively, where $a, b \in R$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^{2} + \mu^{2} = 2$. Then $\frac{a}{b}$ is equal to:

Let $X$ be a random variable with the following probability distribution:

$x$	$-2$	$-1$	$3$	$4$	$6$
$P(X=x)$	$\frac{1}{5}$	$a$	$\frac{1}{3}$	$\frac{1}{5}$	$b$

If the mean of $X$ is $2.3$ and the variance of $X$ is $\sigma^{2}$,then $100 \sigma^{2}$ is equal to:

$A$ random variable $X$ has the probability distribution as shown below. For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$,the probability $P(E \cup F)$ is:

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$