A random variable X takes the values – 2, -1, 1 and 2 with probabilities 1−a4,1+2a4,1−2a4 and 1+a4 respectively then

0≤1−a4≤1⇒0≤1−a≤4⇒0−1≤−a≤3⇒-3≤a≤10≤1+2a4≤1⇒0≤1+2a≤4⇒−1≤2a≤3⇒−12≤a≤32 0≤1−2a4≤1⇒0≤1−2a≤4⇒−1≤−2a≤3⇒-32≤a≤120≤1+a4≤10≤1+a≤40≤a≤3therefore a∈−12,12

A random variable X takes the values – 2, -1, 1 and 2 with probabilities 1−a4,1+2a4,1−2a4 and 1+a4 respectively then

0≤1−a4≤1⇒0≤1−a≤4⇒0−1≤−a≤3⇒-3≤a≤10≤1+2a4≤1⇒0≤1+2a≤4⇒−1≤2a≤3⇒−12≤a≤32 0≤1−2a4≤1⇒0≤1−2a≤4⇒−1≤−2a≤3⇒-32≤a≤120≤1+a4≤10≤1+a≤40≤a≤3therefore a∈−12,12

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A random variable X takes the values – 2, -1, 1 and 2 with probabilities $\frac{1 - a}{4}, \frac{1 + 2 a}{4}, \frac{1 - 2 a}{4}$ and $\frac{1 + a}{4}$ respectively then

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$a c a n h a v e a n y r e a l v a l u e$

$- \frac{1}{2} \leq a \leq \frac{1}{2}$

$- 1 \leq a \leq 1$

$\frac{1}{4} \leq a \leq \frac{1}{3}$

answer is B.

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Detailed Solution

$\begin{array}{l} 0 \leq \frac{1 - a}{4} \leq 1 \\ \Rightarrow 0 \leq 1 - a \leq 4 \\ \Rightarrow 0 - 1 \leq - a \leq 3 \\ \Rightarrow - 3 \leq a \leq 1 \end{array}$

$\begin{array}{l} 0 \leq \frac{1 + 2 a}{4} \leq 1 \\ \Rightarrow 0 \leq 1 + 2 a \leq 4 \\ \Rightarrow - 1 \leq 2 a \leq 3 \\ \Rightarrow \frac{- 1}{2} \leq a \leq \frac{3}{2} \end{array}$

$\begin{array}{l} 0 \leq \frac{1 - 2 a}{4} \leq 1 \\ \Rightarrow 0 \leq 1 - 2 a \leq 4 \\ \Rightarrow - 1 \leq - 2 a \leq 3 \\ \Rightarrow - \frac{3}{2} \leq a \leq \frac{1}{2} \end{array}$

$\begin{array}{l} 0 \leq \frac{1 + a}{4} \leq 1 \\ 0 \leq 1 + a \leq 4 \\ 0 \leq a \leq 3 \\ therefore a \in [\frac{- 1}{2}, \frac{1}{2}] \end{array}$