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Home  >>  CBSE XII  >>  Math  >>  Probability
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The random variable X has a probability distribution P(X) of the following form, where k is some number : \[ P(X) = \left\{ \begin{array}{l l} k & \quad \text{if $x$ = 0}\\ 2k & \quad \text{if $x$ = 1} \\ 3k & \quad \text{if $x$ = 2} \\ 0 & \quad \text{otherwise} \end{array} \right. \] (a) Determine the value of k. (b) Find \(P (X \lt 2), P (X \leq 2), P(X \geq 2).\)

$\begin{array}{1 1} (a) \large \frac{1}{6} \quad \normalsize(b) \; \large\frac{1}{6}, 1, \large\frac{1}{6} \\(b) \large \frac{1}{6} \quad \normalsize(b) \; \large\frac{1}{2}, 1, \large\frac{1}{2} \\(c) \large \frac{1}{3} \quad \normalsize(b) \; \large\frac{1}{3}, 1, \large\frac{1}{6} \\ (d) \large \frac{1}{3} \quad \normalsize(b) \; \large\frac{1}{3}, 1, \large\frac{1}{3}\end{array} $

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Toolbox:
  • To check if a given distribution is a probability distribution of random variable, the sum of the individual probabilties should add up to 1 (i.e, $\sum P(X_i) = 1$). Also 0 $\lt$ P(X) $\leq$ 1.
(a) Since the sum of a probability distribution is always 1 we can $\sum P(X_i) = 1$ and solve for k.
$\Rightarrow k+2k+3k = 1 \rightarrow 6k = 1 \rightarrow k = \large\frac{1}{6}$
P (X $\lt$ 2) = $\sum P(X_i) = 1$ for $ i < 2 = k+2k =3k = 3\times\large\frac{1}{6} = \frac{1}{2}$
P (X $\leq$ 2) = $\sum P(X_i) = 1$ for $ i \leq 2 = k+2k+3k =6k = 6\times\large\frac{6}{6} $$=1$
P (X $\gt$ 2) = P(2) + P(otherwise) = P(2) $= 3k = 3 \times\large\frac{1}{6} = \frac{1}{2}$
answered Jun 20, 2013 by balaji.thirumalai
 

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