Let X be a discrete random variable whose probability distribution is defined as :
\(P(X = x) = \begin{cases} 0.5, & \text{if } x = 0\\ k(x + 1), & \text{if } x = 1 \text{ or } 2\\ k(6 - x), & \text{if } x = 3 \text{ or } 4\\ 0, & \text{otherwise}\end{cases}\)
The value of k is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{20}\)
The probabilities of all values of X add up to 1:
\(P(0) = 0.5\)
\(P(1) = 2k,\quad P(2) = 3k\)
\(P(3) = 3k,\quad P(4) = 2k\)
\(0.5 + 2k + 3k + 3k + 2k = 1\)
\(10k = 0.5\)
\( \displaystyle k = \frac{1}{20} \)