Let \( X \) be a discrete random variable whose probability distribution is defined as follows;
\( P(X=x)=\begin{cases}k(x+1), & x=1,2,3,4\\ 2kx\quad, & x=5,6,7\\ 0\quad, & otherwise\end{cases} \)
where \( k \) is constant , then \( P(X=5) \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{1}{5} \)
The probabilities add up to 1:
\( k(2+3+4+5)+2k(5+6+7)=1 \)
\( 14k+36k=1 \)
\( \displaystyle k=\frac{1}{50} \)
Then
\( P(X=5)=2k\times 5=10k \)
\( \displaystyle =\frac{10}{50}=\frac{1}{5} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{1}{10} \) — Incorrect. The value is \( \frac{1}{5} \), not \( \frac{1}{10} \).
Option (2) \( \displaystyle\frac{1}{50} \) — Incorrect. This is the value of k, not of P(X = 5).
Option (4) \( \displaystyle\frac{1}{20} \) — Incorrect. The value is \( \frac{1}{5} \), not \( \frac{1}{20} \).