Let x be a random variable which assumes values \( x_1,x_2,x_3,x_4 \), such that \( 2P(x=x_1)=3P(x=x_2)=P(x=x_3)=5P(x=x_4) \), then the value of \( P(x=x_4) \) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{6}{61} \)
Let the common value be \( k \), so that
\( \displaystyle P(x_1)=\frac{k}{2},\quad P(x_2)=\frac{k}{3},\quad P(x_3)=k,\quad P(x_4)=\frac{k}{5} \)
The probabilities add up to 1:
\( \displaystyle k\left(\frac{1}{2}+\frac{1}{3}+1+\frac{1}{5}\right)=1 \)
\( \displaystyle \frac{61k}{30}=1 \)
\( \displaystyle k=\frac{30}{61} \)
Hence \( P(x_4)=\frac{k}{5}=\frac{6}{61} \).