$A$ random variable $X$ has the following probability distribution:
$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 $F(4) = $

  • A
    $\frac{3}{10}$
  • B
    $\frac{1}{10}$
  • C
    $\frac{7}{10}$
  • D
    $\frac{4}{5}$

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The p.m.f of a random variable $X$ is given by $P(X) = \frac{2x}{n(n+1)}$ for $x = 1, 2, 3, \ldots, n$,and $0$ otherwise. Then $E(X) = $

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