Question

A random variable X has the following probability distribution
$\begin{array}{cccc}X& P(X=x)& X& P(X=x)\\ 0& \lambda & 5& 11\lambda \\ 1& 3\lambda & 6& 13\lambda \\ 2& 5\lambda & 7& 15\lambda \\ 3& 7\lambda & 8& 17\lambda \\ 4& 9\lambda & & \end{array}$
then, $\lambda$ is equal to

• A. $\frac{1}{81}$
• B. $\frac{2}{81}$
• C. $\frac{5}{81}$
• D. $\frac{7}{81}$
  

Answer 1

The correct option is A $\frac{1}{81}$

${\sum }_{x=0}^{8}P\left(X\right)=1\Rightarrow P\left(0\right)+P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+P\left(6\right)+P\left(7\right)+P\left(8\right)=1=\lambda +3\lambda +5\lambda +7\lambda +9\lambda +11\lambda +13\lambda +15\lambda +17\lambda =1\Rightarrow \frac{9}{2}(\lambda +17\lambda )=1\Rightarrow 81\lambda =1\therefore \lambda =\frac{1}{81}$