Let X be random variable which assumes $x_1, x_2, x_3, x_4$ such that $2P(X=x_1)= 3P(X=x_2)=P(X=x_3)=5P(X=x_4)$, then the probability distribution of X is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) **
|
$X$ |
$x_1$ |
$x_2$ |
$x_3$ |
$x_4$ |
|
$P(X)$ |
15/61 |
10/61 |
30/61 |
6/61 |
Given: $2P(X=x_1)=3P(X=x_2)=P(X=x_3)=5P(X=x_4)$
Let the common value be $k$.
$2P(X=x_1)=k \Rightarrow P(X=x_1)=\frac{k}{2}$
$3P(X=x_2)=k \Rightarrow P(X=x_2)=\frac{k}{3}$
$P(X=x_3)=k$
$5P(X=x_4)=k \Rightarrow P(X=x_4)=\frac{k}{5}$
Total probability = $1$
$\frac{k}{2}+\frac{k}{3}+k+\frac{k}{5}=1$
$k\left(\frac{15+10+30+6}{30}\right)=1$
$k\left(\frac{61}{30}\right)=1$
$k=\frac{30}{61}$
Thus,
$P(X=x_1)=\frac{k}{2}=\frac{15}{61}$
$P(X=x_2)=\frac{k}{3}=\frac{10}{61}$
$P(X=x_3)=k=\frac{30}{61}$
$P(X=x_4)=\frac{k}{5}=\frac{6}{61}$
The probability distribution is: $(\frac{15}{61},\frac{10}{61},\frac{30}{61},\frac{6}{61})$.