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(A) We know that the sum of all probabilities in a probability distribution is $1$,i.e.,$\sum P(X=x) = 1$.Given the distribution:$k^2 + 2k + k + 2k +

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$\frac{35}{36}$

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(A) We know that the sum of all probabilities in a probability distribution is $1$,i.e.,$\sum P(X=x) = 1$.Given the distribution:$k^2 + 2k + k + 2k + 5k^2 = 1$$6k^2 + 5k - 1 = 0$$(6k - 1)(k + 1) = 0$Since $P(X) \geq 0$,we must have $k > 0$,so $k = \frac{1}{6}$.Now,we need to find $P(X \geq 2)$:$P(X \geq 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5)$$P(X \geq 2) = 2k + k + 2k + 5k^2 = 5k + 5k^2$Substituting $k = \frac{1}{6}$:$P(X \geq 2) = 5(\frac{1}{6}) + 5(\frac{1}{6})^2$$P(X \geq 2) = \frac{5}{6} + \frac{5}{36} = \frac{30 + 5}{36} = \frac{35}{36}$.

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

$A$ random variable $X$ has the following probability distribution:
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| $P(X)$ | $k^2$ | $2k$ | $k$ | $2k$ | $5k^2$ |
Then $P(X \geq 2)$ is equal to:

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For the probability distribution given below,find $\operatorname{Var}(X)$.
$X$ $5$ $6$ $7$ $8$ $9$ $10$ $11$
$P(X=x)$ $0.07$ $0.2$ $0.3$ $k$ $0.07$ $0.04$ $0.02$

$A$ person throws two fair dice. He wins $Rs.\, 15$ for throwing a doublet (same numbers on the two dice),wins $Rs.\, 12$ when the throw results in the sum of $9$,and loses $Rs.\, 6$ for any other outcome on the throw. Then the expected gain/loss (in $Rs.$) of the person is

The p.d.f. of a continuous random variable $X$ is $f(x)=\begin{cases} \frac{x^2}{18} & \text{if } -3 < x < 3 \\ 0 & \text{otherwise} \end{cases}$. Then $P[|X| < 2]=$

The probability mass function of a random variable $X$ is given by $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x = 0, 1, 2, 3, 4, 5$ and $0$ otherwise. Then,$P(X \leq 2)$ is equal to:

The probability distribution of a random variable $X$ is given below:
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$

Then, $P(0 < X < 4)$ is equal to:

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$A$ random variable $X$ has the following probability distribution:| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ ||---|---|---|---|---|---|| $P(X)$ | $k^2$ | $2k$ | $k$ | $2k$ | $5k^2$ |Then $P(X \geq 2)$ is equal to: