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(B) For a probability distribution,the sum of all probabilities must be $1$.$\sum P(X) = K + 2K + 2K + 3K + K = 9K = 1 \Rightarrow K = \frac{1}{9}$.We

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$30$

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(B) For a probability distribution,the sum of all probabilities must be $1$.$\sum P(X) = K + 2K + 2K + 3K + K = 9K = 1 \Rightarrow K = \frac{1}{9}$.We need to find $p = P(1 By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.Here,$A = \{2, 3\}$ and $B = \{1, 2\}$.$A \cap B = \{2\}$.So,$p = \frac{P(X=2)}{P(X=1) + P(X=2)} = \frac{2K}{K + 2K} = \frac{2K}{3K} = \frac{2}{3}$.Given $5p = \lambda K$,we substitute the values:$5 	imes \left(\frac{2}{3}ight) = \lambda 	imes \left(\frac{1}{9}ight)$.$\frac{10}{3} = \frac{\lambda}{9}$.$\lambda = \frac{10 	imes 9}{3} = 30$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$K$	$2K$	$2K$	$3K$	$K$

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

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

$X=x$	$-3$	$0$	$1$	$\alpha$
$P(X=x)$	$\frac{1}{4}$	$K$	$\frac{1}{4}$	$\frac{1}{3}$

The probability distribution of random variable $X$ is given by:

$X$ $1$ $2$ $3$ $4$ $5$

$P(X)$ $K$ $2K$ $2K$ $3K$ $K$

Let $p=P(1 < X < 4 \mid X < 3)$. If $5p = \lambda K$,then $\lambda$ is equal to .... .

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Then $E(x) = ?$

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The probability distribution of random variable $X$ is given by: $X$ $1$ $2$ $3$ $4$ $5$ $P(X)$ $K$ $2K$ $2K$ $3K$ $K$ Let $p=P(1 < X < 4 \mid X < 3)$. If $5p = \lambda K$,then $\lambda$ is equal to .... .