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(B) For a probability distribution,the sum of all probabilities must be $1$.$\sum P(X) = k + 2k + 4k + 2k + k = 10k = 1 \implies k = \frac{1}{10}$.We

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$\frac{6}{7}$

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(B) For a probability distribution,the sum of all probabilities must be $1$.$\sum P(X) = k + 2k + 4k + 2k + k = 10k = 1 \implies k = \frac{1}{10}$.We need to find the conditional probability $P(1 \le X By the definition of conditional probability,$P(A|B) = \frac{P(A \cap B)}{P(B)}$.Here,$A$ is the event $1 \le X $B$ is the event $X \le 2$,which means $X \in \{0, 1, 2\}$.The intersection $A \cap B$ is the event $X \in \{1, 2\}$.Now,$P(A \cap B) = P(X=1) + P(X=2) = 2k + 4k = 6k = 6 	imes \frac{1}{10} = \frac{6}{10}$.And $P(B) = P(X=0) + P(X=1) + P(X=2) = k + 2k + 4k = 7k = 7 	imes \frac{1}{10} = \frac{7}{10}$.Therefore,$P(A|B) = \frac{6/10}{7/10} = \frac{6}{7}$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$k$	$2k$	$4k$	$2k$	$k$

$A$ random variable $X$ has the following probability distribution:

$X$ $0$ $1$ $2$ $3$ $4$

$P(X)$ $k$ $2k$ $4k$ $2k$ $k$

Then the value of $P(1 \le X < 4 | X \le 2) =$ ?

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$A$ random variable $X$ has the following probability distribution: $X$ $0$ $1$ $2$ $3$ $4$ $P(X)$ $k$ $2k$ $4k$ $2k$ $k$ Then the value of $P(1 \le X < 4 | X \le 2) =$ ?