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(D) For a probability distribution,the sum of probabilities must be $1$:$\frac{1}{5} + A + B = 1 \implies A + B = 1 - \frac{1}{5} = \frac{4}{5}$ (Equa

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$\frac{3}{20}$

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(D) For a probability distribution,the sum of probabilities must be $1$:$\frac{1}{5} + A + B = 1 \implies A + B = 1 - \frac{1}{5} = \frac{4}{5}$ (Equation $1$).The expected value $E(X)$ is given by $\sum X \cdot P(X)$:$E(X) = 30 \cdot (\frac{1}{5}) + 10 \cdot A + (-10) \cdot B = 4$$6 + 10A - 10B = 4$$10A - 10B = -2 \implies 5A - 5B = -1$ (Equation $2$).From Equation $1$,$B = \frac{4}{5} - A$. Substituting this into Equation $2$:$5A - 5(\frac{4}{5} - A) = -1$$5A - 4 + 5A = -1$$10A = 3 \implies A = \frac{3}{10}$.Now,find $B$: $B = \frac{4}{5} - \frac{3}{10} = \frac{8-3}{10} = \frac{5}{10} = \frac{1}{2}$.The value of $AB = (\frac{3}{10}) \cdot (\frac{1}{2}) = \frac{3}{20}$.

$X$	$30$	$10$	$-10$
$P(X)$	$\frac{1}{5}$	$A$	$B$

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$P(x)$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

Outcome	Probability
$\omega_{1}$	$\frac{1}{12}$
$\omega_{2}$	$\frac{1}{12}$
$\omega_{3}$	$\frac{1}{6}$
$\omega_{4}$	$\frac{1}{6}$
$\omega_{5}$	$\frac{1}{6}$
$\omega_{6}$	$\frac{3}{2}$

Let $X$ be a discrete random variable. The probability distribution of $X$ is given below:
$X$ $30$ $10$ $-10$
$P(X)$ $\frac{1}{5}$ $A$ $B$

If $E(X) = 4$,then the value of $AB$ is equal to:

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Let $X$ be a discrete random variable. The probability distribution of $X$ is given below:$X$$30$$10$$-10$$P(X)$$\frac{1}{5}$$A$$B$If $E(X) = 4$,then the value of $AB$ is equal to: