$A$ random variable $X$ has the probability distribution as given below. Let $E = \{X \mid X \text{ is a prime number}\}$ and $F = \{X \mid X < 4\}$,then $P(E \cup F) = $
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline P(X) & K & 2K & K^2 & 2K^2 & 5K^2 & K & K & 2K \\ \hline \end{array}$
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline P(X) & K & 2K & K^2 & 2K^2 & 5K^2 & K & K & 2K \\ \hline \end{array}$
- A$\frac{38}{64}$
- B$\frac{39}{64}$
- C$\frac{42}{64}$
- D$\frac{17}{64}$
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Similar Questions
For the given probability distribution,find $E(X^2)$.$X$ $1$ $2$ $3$ $4$ $P(X)$ $\frac{1}{10}$ $\frac{1}{5}$ $\frac{3}{10}$ $\frac{2}{5}$
| $X$ | $1$ | $2$ | $3$ | $4$ |
| $P(X)$ | $\frac{1}{10}$ | $\frac{1}{5}$ | $\frac{3}{10}$ | $\frac{2}{5}$ |
If the probability distribution of a random variable $X$ is given by the following table, then the mean of $X$ is:
$X = x$ $0$ $2$ $4$ $6$ $8$ $10$ $P(X = x)$ $0$ $k$ $2k$ $5k^2$ $2k^2$ $3k$
| $X = x$ | $0$ | $2$ | $4$ | $6$ | $8$ | $10$ |
|---|---|---|---|---|---|---|
| $P(X = x)$ | $0$ | $k$ | $2k$ | $5k^2$ | $2k^2$ | $3k$ |
If a random variable $X$ follows a Poisson distribution with a mean value of $5$,then the probability that $X < 3$ is:
The random variable $X$ has the following probability distribution:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ is:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ is:
For the following probability distribution, the standard deviation of the random variable $X$ is:
$X: 0, 1, 2$
$P(X): 0.3, 0.4, 0.3$
$X: 0, 1, 2$
$P(X): 0.3, 0.4, 0.3$
DifficultMHT CET 2026
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