13. The probability distribution of a random variable X is given by:

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

If $a>0$, then $P(0<X\le 2)$ is equal to:

The correct option is:
$\frac{7}{16}$

Step 1: Understand the properties of a probability distribution
For any discrete random variable $X$, the sum of all probabilities in its probability distribution must be equal to 1.
Mathematically, this is expressed as:
$\sum P\left(X\right)=1$

From the given table, the values of $X$ are 0, 1, and 2, with corresponding probabilities:
$P(X=0)=1-7a^2$
$P(X=1)=\frac{a}{2}+\frac{1}{4}$
$P(X=2)=a^2$

Step 2: Set up the equation to find $a$
Adding these probabilities together, we get:
$(1-7a^2)+(\frac{a}{2}+\frac{1}{4})+a^2=1$

Subtracting 1 from both sides simplifies the equation to:
$-6a^2+\frac{a}{2}+\frac{1}{4}=0$

To eliminate the denominators, multiply the entire equation by 4:
$-24a^2+2a+1=0$
Multiplying by -1 gives a standard quadratic equation:
$24a^2-2a-1=0$

Step 3: Solve the quadratic equation for $a$
We can factor this quadratic expression by finding two numbers that multiply to $24\times (-1)=-24$ and add up to $-2$. These numbers are $-6$ and $4$.
Split the middle term:
$24a^2-6a+4a-1=0$
Factor by grouping:
$6a(4a-1)+1(4a-1)=0$
$(6a+1)(4a-1)=0$

This yields two possible values for $a$:
$a=-\frac{1}{6}$ or $a=\frac{1}{4}$

Since we are given that $a>0$, we discard the negative value. Therefore, we have:
$a=\frac{1}{4}$

Step 4: Calculate the required probability $P(0<X\le 2)$
The inequality $0<X\le 2$ is satisfied by the discrete values $X=1$ and $X=2$.
Therefore:
$P(0<X\le 2)=P(X=1)+P(X=2)$

Substitute the algebraic expressions for these probabilities:
$P(0<X\le 2)=(\frac{a}{2}+\frac{1}{4})+a^2$

Now, substitute $a=\frac{1}{4}$ into the equation:
$P(0<X\le 2)=(\frac{1/4}{2}+\frac{1}{4})+{\left(\frac{1}{4}\right)}^2$
Simplify each term step-by-step:
$\frac{1/4}{2}=\frac{1}{8}$
${\left(\frac{1}{4}\right)}^2=\frac{1}{16}$

Combine the terms by finding a common denominator of 16:
$P(0<X\le 2)=\frac{1}{8}+\frac{1}{4}+\frac{1}{16}$
$P(0<X\le 2)=\frac{2}{16}+\frac{4}{16}+\frac{1}{16}$
$P(0<X\le 2)=\frac{2+4+1}{16}=\frac{7}{16}$