Question:
For a discrete random variable X, whose probability distribution is defined as :
\(P(x)=\begin{cases} 2k(x+1) ;& x = 0,1 \\ 3kx; & x=2 \\ k(5-x) & x=3 \end{cases}\)
The value of mean will be
For a discrete random variable X, whose probability distribution is defined as :
\(P(x)=\begin{cases} 2k(x+1) ;& x = 0,1 \\ 3kx; & x=2 \\ k(5-x) & x=3 \end{cases}\)
The value of mean will be
\(P(x)=\begin{cases} 2k(x+1) ;& x = 0,1 \\ 3kx; & x=2 \\ k(5-x) & x=3 \end{cases}\)
The value of mean will be
Updated On: Jun 13, 2024
- \(\frac{6}{7}\)
- \(\frac{15}{7}\)
- \(\frac{12}{7}\)
- \(\frac{11}{7}\)
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The Correct Option is D
Solution and Explanation
The correct option is (D) :\(\frac{11}{7}\).
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