The random variable $x$ has a probability distribution $p(x)$ of the form $p(x = r) =\begin{cases}r \cdot k & \text{if } r \leq 2 \\(r-1)k & \text{if } 2 < r \leq 4 \\0 & \text{otherwise}\end{cases}$ where $r \in \mathbb{N} \cup \{0\}$ and $k \in \mathbb{R}$. Then:
(A) $k = \frac{1}{9}$
(B) $P(2 \leq x \leq 3) = \frac{1}{2}$
(C) $P(x = 4) = \frac{1}{3}$
(D) $P(x > 1) = \frac{7}{8}$
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (D) only
(B) $P(2 \leq x \leq 3) = \frac{1}{2}$
(D) $P(x > 1) = \frac{7}{8}$
Possible $r$: $0,1,2,3,4$
$p(0)=0\cdot k=0,\quad p(1)=1\cdot k=k,\quad p(2)=2k$
$p(3)=(3-1)k=2k,\quad p(4)=(4-1)k=3k$
Normalization: $0+k+2k+2k+3k=8k=1 \Rightarrow k=\frac{1}{8}$
Check (A): $\frac{1}{9}$ — Incorrect.
$P(2\le x\le 3)=p(2)+p(3)=2k+2k=4k=\frac{4}{8}=\frac{1}{2}$ — (B) correct.
$P(x=4)=3k=\frac{3}{8}$ — (C) incorrect.
$P(x>1)=p(2)+p(3)+p(4)=2k+2k+3k=7k=\frac{7}{8}$ — (D) correct.
Correct: (B), (D)