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  • Let X be a random variable which assumes values  such that  <img alt="2P\left ( X= x_{1} \right )

Let X be a random variable which assumes values x_{1},x_{2},x_{3},x_{4} such that  2P\left ( X= x_{1} \right )= 3P\left ( X= x_{2} \right )= P\left ( X= x_{3} \right )= 5P\left ( X= x_{4} \right ).Find the probability distribution of X.

 

 

 

 
 
 
 
 

Answers (1)

given:
           x_{1},\, x_{2},\, x_{3},\, x_{4 }  random variables
2P\left ( X= x_{1} \right )= 3P\left ( X= x_{2} \right )= P\left ( X= x_{3} \right )= 5P\left ( X= x_{4} \right )
Let   P\left ( x= x_{3} \right )= x
        P\left ( x= x_{2} \right )= \frac{x}{3}\: \: \: \: \: P\left ( x= x_{1} \right )= \frac{x}{2}
       P\left ( x= x_{4} \right )= \frac{x}{5}
\sum_{x=1}^{4}P\left ( x_{1} \right )= 1
P\left ( x_{1} \right )+P\left ( x_{2} \right )+P\left ( x_{3} \right )+P\left ( x_{4} \right )= 1
\frac{x}{2}+\frac{x}{3}+x+\frac{x}{5}= 1
x= \frac{30}{61}\: \: \: \: \: P\left (x _{3} \right )= \frac{30}{61}
P\left ( x_{1} \right )= \frac{30}{61\times 2}= \frac{15}{61}\: \: \: P\left ( x_{2} \right )= \frac{\not{30}}{61}\times \frac{1}{3}= \frac{10}{61}
P\left ( x_{4} \right )= \frac{\not{30}}{61}\times \frac{1}{\not{5}}\Rightarrow \frac{6}{61}
so the probability distribution function will be
X\: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\:\: 1\:\: \:\: 2\:\:\: \: \: 3\:\:\: \: \: 4
P\left ( x= xi \right )\:\: \: \frac{15}{61}\: \: \frac{10}{61}\: \: \frac{30}{61}\: \: \frac{6}{61}???????

Posted by

Ravindra Pindel

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