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Q)

A random variable $X$ has the following probability distribution.\[\]$\begin{array} {llllllll} \textbf{X:}& 0& 1& 2& 3& 4& 5 \\ \textbf{P(X=x):}& 1/4& 2a &3a& 4a& 5a& 1/4 \end{array}$\[\]Then $p(1\leq x\leq 4)$ is

\[\begin{array}{1 1}(1)\frac{10}{21}&(2)\frac{2}{7}\\(3)\frac{1}{14}&(4)\frac{1}{2}\end{array}\]

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A)
$\begin{array} {llllllll} \textbf{X:}& 0& 1& 2& 3& 4& 5 \\ \textbf{P(X=x):}& 1/4& 2a &3a& 4a& 5a& 1/4 \end{array}$
$\large\frac{1}{4} $$+2a+3a+4a+5a +\large\frac{1}{4}$$=1$
$\large\frac{1}{2} $$+14 a=1$
$14 a =\large\frac{1}{2}$
$a= \large\frac{1}{28}$
$P(1 \leq x \leq 4) =P(x=1) +P(x=2)$
$P(x=3) +P(x=4)$
$\qquad= 2a+3a+4a+5a$
$\qquad=14 a$
$\qquad= 14 \times \large\frac{1}{28}$
$\qquad= \large\frac{1}{2}$
Hence 4 is the correct answer.
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