The probability distribution of a random variable X is given by

X	0	1	2
P(X)	$1 - 7a^2$	$\frac{1}{2}a + \frac{1}{4}$	$a^2$

If $a > 0$, then $P(0 < x \leq 2)$ is equal to

The probability $P(0 < X \leq 2)$ represents values where $X$ is greater than 0 and less than or equal to 2.

From the table, this includes $X = 1$ and $X = 2$ (but not $X = 0$).

$P(0 < X \leq 2) = P(X=1) + P(X=2)$

In any probability distribution, all probabilities sum to 1.

$P(X=0) + P(X=1) + P(X=2) = 1$

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

$1 - 7a^2 + \frac{1}{2}a + \frac{1}{4} + a^2 = 1$

$\frac{5}{4} - 6a^2 + \frac{1}{2}a = 1$

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

Multiplying by 4:

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

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

$a = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(24)(-1)}}{2(24)}$

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

$a = \frac{2 \pm \sqrt{100}}{48}$

$a = \frac{2 \pm 10}{48}$

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

Since $a > 0$, we have $a = \frac{1}{4}$

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

$P(X=1) = \frac{1}{8} + \frac{2}{8}$

$P(X=1) = \frac{3}{8}$

$P(X=2) = (\frac{1}{4})^2$

$P(X=2) = \frac{1}{16}$

$P(0 < X \leq 2) = P(X=1) + P(X=2)$

$P(0 < X \leq 2) = \frac{3}{8} + \frac{1}{16}$

$P(0 < X \leq 2) = \frac{6}{16} + \frac{1}{16}$

$P(0 < X \leq 2) = \frac{7}{16}$