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Home  >>  CBSE XII  >>  Math  >>  Probability
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Given the following probability distribution, $\begin{matrix} \text{X} & 0 &1 &2 &3&4&5&6&7\\ \text{P(X)}&0&k&2k&2k&3k&k^2&2k^2&7k^2+k \end{matrix}$ \[\]determine $\;(i) k\quad (ii) P(X \lt 3)\quad (iii) P(X \gt 6)\quad (iv) P(0 \lt X \lt 3)$

$\begin{array}{1 1}(i) .10, (ii) .30, (iii) .17, (iv) .30 \\ (i) .10, (ii) .20, (iii) .07, (iv) .30 \\(i) .05, (ii) .30, (iii) .17, (iv) .20 \\ (i) .05, (ii) .30, (iii) .07, (iv) .20 \end{array} $

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1 Answer

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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.
(i) Since the sum of a probability distribution is always 1 we can $\sum P(X_i) = 1$ and solve for k.
$\Rightarrow 0+k+2k+2k+3k+k^2+2k^2+7k^2+k = 1$
$\Rightarrow 10k^2+9-1=0$
Solving, we get $k=-1$ or $k=\large\frac{1}{10}$. Since we cannot have negative probabilities, k = $\large\frac{1}{10}$
P (X $\lt$ 3) = $\sum P(X_i) = 1$ for $ i < 3 = 0 + k + 2k = 3k = 3\times\large\frac{1}{10} = \frac{3}{10}$
P (X $\gt$ 6) = $\sum P(X_i) = 1$ for $ i > 3 = 7k^2+k = 7\times\large\frac{1}{10}^2+\frac{1}{10} = \frac{17}{100}$
P (0 $\lt$ X $\lt$ 3) = $\sum P(X_i) = 1$ for $0< i < 3 = k + 2k = 3k = 3\times\large\frac{1}{10} = \frac{3}{10}$
answered Jun 20, 2013 by balaji.thirumalai
 

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