The random variable X has a probability distribution P(X) of the following form, where ‘k’ is some number. 

P(X=x)=\begin{cases}k,&i\!f\: \: x=0\\ 2k,&i\!f\: \: x=1\\ 3k,&i\!f\: \: x=3\\0,&otherwise \end{cases}

Determine the value of ‘k’. 

 

 

 

 
 
 
 
 

Answers (1)

As we know sum of all probablities in a probablity distribution is 1.

i.e   \sum p=1

\therefore k+2k+3k+0=1

\Rightarrow 6k=1

\Rightarrow k=\frac{1}{6}

Which is the required value

 

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