The probability density for finding a particle at position $x$is

$p\left(x\right)=\left\{\begin{array}{l}\frac{a}{\left(1-x\right)}\\ b\left(1-x\right)\end{array}\right.\frac{-1mm\le x<0mm}{0mm\le x\le 1mm}$

and zero elsewhere

In a quadrium world the trajectory a particle could not be exactly predicted . The probability $p\left(x\right)$of finding the particle a certain location in wave is determined by the calculating the expectation value of the wave function localid="1648836367063" $$\begin{gathered} p\left(x\right)oftheparticle \\ p\left(x\right)=\int v\left(x\right)dx{\int }_{}^3 \end{gathered}$$

The wave function is continuous. We can apply the boundary conditions, so the wave function at boundaries is same.

$$\begin{gathered} \underset{x\to \infty }{\lim }p\left(x\right)=\underset{x\to \infty }{\lim }p\left(x\right) \\ \underset{x\to \infty }{\lim }-\frac{a}{1-x}-\underset{x\to \infty }{\lim }b(1-x) \end{gathered}$$

Therefore, we can conclude that the relation between a and b is a=b

${\int }_{-\infty }^{\infty }p\left(x\right)dx=1$

Now substitute the value of P(x) given in the problem in the above equation

$$\begin{gathered} {\int }_{-1}^0\frac{a}{1-x}dx+{\int }_0^1{\left(-4cx+4cx\right)}^{2}dx=1 \\ -aIn{(1-x)}_{-1}^0+b{\left(x-\frac{x2}{2}\right)}_0^1=1 \\ aIn2+b/2=1 \end{gathered}$$

substitute b for a

$$\begin{gathered} aLn2+a/2=1 \\ a=\frac{2}{2In2+1} \\ =0.84 \end{gathered}$$

Therefore, the numerical value of a and b is 0.84

$$\begin{gathered} p\left(x\right)={\int }_{-1}^0\frac{a}{1-x}dx \\ =a{\left\{-In\left(1-x\right)\right\}}_{-1}^0 \\ =aIn2 \end{gathered}$$

Substitute 0.84 for a

=(0.84)In2

=0.581

Therefore the probability of finding the particle in the left side of the origin is 58.1%