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(B) For a probability distribution,the sum of probabilities must be $1$:$\frac{1+p}{5} + \frac{2-2p}{5} + \frac{2-p}{5} + \frac{2p}{5} = 1$$\frac{1+p+

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$6$

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(B) For a probability distribution,the sum of probabilities must be $1$:$\frac{1+p}{5} + \frac{2-2p}{5} + \frac{2-p}{5} + \frac{2p}{5} = 1$$\frac{1+p+2-2p+2-p+2p}{5} = 1$$\frac{5}{5} = 1$. This is always true for any $p$.Since $P(X=x) \ge 0$ for all $x$,we have:$1+p \ge 0 \implies p \ge -1$$2-2p \ge 0 \implies p \le 1$$2-p \ge 0 \implies p \le 2$$2p \ge 0 \implies p \ge 0$Combining these,$0 \le p \le 1$. The minimum value of $p$ is $0$.Now,calculate $E(X) = \sum x P(X=x)$:$E(X) = 0 \cdot \frac{1+p}{5} + 1 \cdot \frac{2-2p}{5} + 2 \cdot \frac{2-p}{5} + 3 \cdot \frac{2p}{5}$$E(X) = \frac{2-2p + 4-2p + 6p}{5} = \frac{6+2p}{5}$For $p=0$,$E(X) = \frac{6}{5}$.Therefore,$5 E(X) = 5 \cdot \frac{6}{5} = 6$.

$X = k$	$-2$	$-1$	$0$	$1$	$2$
$P(X = k)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$

The following is the probability distribution of $X$:

$X$ $0$ $1$ $2$ $3$

$P(X=x)$ $\frac{1+p}{5}$ $\frac{2-2p}{5}$ $\frac{2-p}{5}$ $\frac{2p}{5}$

For a minimum value of $p$,the value of $5 E(X)$ is:

Similar Questions

The variance of the random variable $X$ having the following distribution is:
$X = k$ $-2$ $-1$ $0$ $1$ $2$
$P(X = k)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$

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The following is the probability distribution of $X$: $X$ $0$ $1$ $2$ $3$ $P(X=x)$ $\frac{1+p}{5}$ $\frac{2-2p}{5}$ $\frac{2-p}{5}$ $\frac{2p}{5}$ For a minimum value of $p$,the value of $5 E(X)$ is: