Let X be a distance random variable whose probability distribution is defined as :

$P(X=x)= \left\{\begin{matrix}0.5 & ,if & x=0\\k(x+1) & ,if & x=1\, or \, 2\\k(6-x) & ,if & x=3\, or\, 4\\0 & , & otherwise\end{matrix}\right.$

The, value of k is :

$\frac{1}{20}$

The correct answer is Option (2) → $\frac{1}{20}$

Since this is a probability distribution, the sum of all probabilities must be equal to 1.

Now write the probabilities for each value of x:

• When x=0, probability = 0.5
• When x=1, probability = 2k
• When x=2, probability = 3k
• When x=3, probability = 3k
• When x=4, probability = 2k

Now add all probabilities:

0.5 + 2k + 3k + 3k + 2k = 1

0.5 + 10k = 1

10k = 0.5

k = 0.5 ÷ 10 = 1/20