The probability distribution of a random variable $X$ is given below:
$X=x$$0$$1$$2$$3$$4$$5$$6$$7$
$P(X=x)$$0$$K$$2K$$2K$$3K$$K^2$$2K^2$$7K^2+K$

Then,$P(0 < X < 5)$ is equal to:

  • A
    $\frac{1}{10}$
  • B
    $\frac{3}{10}$
  • C
    $\frac{8}{10}$
  • D
    $\frac{7}{10}$

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Similar Questions

In a meeting,$70 \%$ of the members favour and $30 \%$ oppose a certain proposal. $A$ member is selected at random and we take $X=0$ if he opposed,and $X=1$ if he is in favour. Find $E(X)$ and $\text{Var}(X)$.

$A$ 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) = $

Given the probability density function: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$ The probability $P\left(\frac{1}{4} < X < \frac{1}{3}\right)$ is given by: $P\left(\frac{1}{4} < X < \frac{1}{3}\right) = \int_{1/4}^{1/3} 3(1 - 2x^2) \, dx$

The p.d.f. of a continuous random variable $X$ is given by $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$ and $f(x) = 0$ otherwise. Then $P[|x| < 1] = $

$A$ random variable $X$ has the following probability distribution:
$x$$0$$1$$2$$3$$4$$5$$6$$7$$8$
$P(X=x)$$k$$2k$$3k$$4k$$4k$$3k$$2k$$k$$k$

Then $P(3 < X \leq 6) = $

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