The probability distribution of a random vari | vedclass

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(C) For a probability distribution,the sum of all probabilities must be equal to $1$,i.e.,$\sum P(X=x) = 1$.Summing the given probabilities:$0 + K + 2

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$\frac{8}{10}$

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(C) For a probability distribution,the sum of all probabilities must be equal to $1$,i.e.,$\sum P(X=x) = 1$.Summing the given probabilities:$0 + K + 2K + 2K + 3K + K^2 + 2K^2 + (7K^2 + K) = 1$Combining like terms:$(K + 2K + 2K + 3K + K) + (K^2 + 2K^2 + 7K^2) = 1$$9K + 10K^2 = 1$$10K^2 + 9K - 1 = 0$Factoring the quadratic equation:$10K^2 + 10K - K - 1 = 0$$10K(K + 1) - 1(K + 1) = 0$$(10K - 1)(K + 1) = 0$This gives $K = \frac{1}{10}$ or $K = -1$. Since probability cannot be negative,$K$ must be positive,so $K = \frac{1}{10}$.We need to find $P(0 $P(X=1) + P(X=2) + P(X=3) + P(X=4)$$= K + 2K + 2K + 3K = 8K$Substituting $K = \frac{1}{10}$:$P(0 < X < 5) = 8 	imes \frac{1}{10} = \frac{8}{10}$.

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

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,$P(0 < X < 5)$ is equal to:

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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 < 5)$ is equal to: