(B) Given,$E = \{x \text{ is a prime number}\} = \{2, 3, 5, 7\}$.$P(E) = P(2) + P(3) + P(5) + P(7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.Given,$F = \{x

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(B) Given,$E = \{x \text{ is a prime number}\} = \{2, 3, 5, 7\}$.$P(E) = P(2) + P(3) + P(5) + P(7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.Given,$F = \{x $P(F) = P(1) + P(2) + P(3) = 0.15 + 0.23 + 0.12 = 0.50$.The intersection $E \cap F = \{x \text{ is prime and } x $P(E \cap F) = P(2) + P(3) = 0.23 + 0.12 = 0.35$.Using the formula $P(E \cup F) = P(E) + P(F) - P(E \cap F)$:$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

Class 12 Mathematics Probability Probability distribution

Difficult

$A$ random variable $X$ has the following probability distribution:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is:

MHT CET 2008 All MHT CET

A
$0.50$
B
$0.77$
C
$0.35$
D
$0.87$

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$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

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Let the mean and the standard deviation of the probability distribution be $\mu$ and $\sigma$,respectively. If $\sigma - \mu = 2$,then $\sigma + \mu$ is equal to:

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