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Home  >>  CBSE XII  >>  Math  >>  Probability
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An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable ?

$\begin{array}{1 1} \text{Possible values of X = 0, 1. X is not a random variable.} \\\text{Possible values of X = 0, 1, 2. X is not a random variable.} \\\text{Possible values of X = 0, 1, 2. X is a random variable.} \\ \text{Possible values of X = 0,1. X is a random variable.}\end{array} $

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Given that an urn has 5 Red balls and 2 Black balls.
The number of ways in which two balls can be reprsented are {(RR), (RB), (BR), (BB)}
Let X represent the number of black balls. Possible values of X are: X (RR) = 0, X (RB) = 1, X(BR = 1) and X (BB) = 2.
Therefore the possible values of X are 0, 1 and 2.
P (X = 0) = $\large\frac{1}{4}$
P (X = 1) = $\large\frac{2}{4}$
P (X =2) = $\large\frac{1}{4}$
$\sum X_i = P(X=0)+P(X=1)+P(X=2) = \large \frac{1}{4} + \frac{2}{4} + \frac{1}{4}$ = 1.
Therefore X is a random variable.

 

answered Jun 19, 2013 by balaji.thirumalai
 

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