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 :
Answer & explanation
Correct answer: option 2
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