(C) The sum of all probabilities in a distribution must be $1$:$\Sigma P(X = x) = 0.15 + 0.23 + k + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 1$$0.88 + k = 1

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(C) The sum of all probabilities in a distribution must be $1$:$\Sigma P(X = x) = 0.15 + 0.23 + k + 0.10 + 0.20 + 0.08 + 0.07 + 0.05 = 1$$0.88 + k = 1$$k = 0.12$The event $E$ consists of prime numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8\}$,so $E = \{2, 3, 5, 7\}$.The event $F$ consists of values less than $4$,so $F = \{1, 2, 3\}$.The union $E \cup F = \{1, 2, 3, 5, 7\}$.The probability $P(E \cup F) = P(X=1) + P(X=2) + P(X=3) + P(X=5) + P(X=7)$$P(E \cup F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77$.

Class 12 Mathematics Probability Probability distribution

Medium

$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X = x)$ $0.15$ $0.23$ $k$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

For the events $E = \{x : x \text{ is a prime number}\}$ and $F = \{x : x < 4\}$,then $P(E \cup F) = $

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A
$0.57$
B
$0.87$
C
$0.77$
D
$0.35$

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If a random variable $x$ has the probability distribution as follows:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
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Then $P(3 < x \leq 6)$ is equal to:

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$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$k$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

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

$A$ random variable $X$ has the following probability distribution:$X = x$$1$$2$$3$$4$$5$$6$$7$$8$$P(X = x)$$0.15$$0.23$$k$$0.10$$0.20$$0.08$$0.07$$0.05$For the events $E = \{x : x \text{ is a prime number}\}$ and $F = \{x : x < 4\}$,then $P(E \cup F) = $

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$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X = x)$ $0.15$ $0.23$ $k$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

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If a random variable $x$ has the probability distribution as follows:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(x)$ $0$ $2k$ $k$ $3k$ $2k^2$ $2k$ $k^2+k$ $7k^2$

Then $P(3 < x \leq 6)$ is equal to: