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Let X Be a Random Variable Which Assumes Values X1, X2, X3, X4 Such that 2p (X = X1) = 3p (X = X2) = P (X = X3) = 5 P (X = X4). Find the Probability Distribution of X. - Mathematics

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Question

Let X be a random variable which assumes values x1, x2, x3, x4 such that 2P (X = x1) = 3P(X = x2) = P (X = x3) = 5 P (X = x4). Find the probability distribution of X.                                                                                                                                                                                 

Sum

Solution

Let P (X = x3) = k. Then,
P (X = x1) = \[\frac{k}{2}\]
P (X = x2) = \[\frac{k}{3}\]
P (X = x4) = \[\frac{k}{5}\]

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = x1) + P (X = x2) + P (X = x3) + P (X = x4) = 1

\[\Rightarrow \frac{k}{2} + \frac{k}{3} + k + \frac{k}{5} = 1\]
\[ \Rightarrow \frac{15k + 10k + 30k + 6k}{30} = 1\]
\[ \Rightarrow \frac{61k}{30} = 1\]
\[ \Rightarrow k = \frac{30}{61}\]

Now,

xi pi
x1 \[\frac{k}{2}\] = \[\frac{15}{61}\]
x2
\[\frac{k}{3}\] = \[\frac{10}{61}\]
 
x3 k = \[\frac{30}{61}\]
x4
\[\frac{k}{5}\] = \[\frac{6}{61}\]
 
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Chapter 32: Mean and Variance of a Random Variable - Exercise 32.1 [Page 14]

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RD Sharma Mathematics [English] Class 12
Chapter 32 Mean and Variance of a Random Variable
Exercise 32.1 | Q 5 | Page 14

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