The probability distribution of a random variable X is given by X 0 1…

The correct answer is Option (3) → $\frac{7}{16}$

$P(\text{X}=0)=1-7a^2$

$P(\text{X}=1)=\frac{1}{2}a+\frac{1}{4}$

$P(\text{X}=2)=a^2$

$(1-7a^2)+\left(\frac{1}{2}a+\frac{1}{4}\right)+a^2=1$

$-6a^2+\frac{1}{2}a+\frac{1}{4}=0$

$-24a^2+2a+1=0$

$24a^2-2a-1=0$

$a=\frac{2\pm\sqrt{4+96}}{48}=\frac{2\pm10}{48}$

$a=\frac{12}{48}=\frac{1}{4}$ or $a=-\frac{8}{48}=-\frac{1}{6}$

$a=\frac{1}{4}$

$P(0<\text{X}\le2)=P(\text{X}=1)+P(\text{X}=2)$

$P(0<\text{X}\le2)=\left(\frac{1}{2}\cdot\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{4}\right)^2$

$P(0<\text{X}\le2)=\frac{1}{8}+\frac{1}{4}+\frac{1}{16}$

$P(0<\text{X}\le2)=\frac{7}{16}$