91Ó°ÊÓ

Sub part (a) step1:

For the probability interpretation of $\mathrm{ψ}\left(x\right)$to make sense the wave function must satisfy the following equation

${∫}_{-∞}^{+∞}{\left|\mathrm{ψ}\left(x\right)\right|}^{2}dx=1$

The above integrals is expanded as follows${∫}_{-∞}^0{\left|\mathrm{ψ}\left(x\right)\right|}^{2}dx+{∫}_0^1{\left|\mathrm{ψ}\left(x\right)\right|}^{2}dx+{∫}_1^{+∞}{\left|\mathrm{ψ}\left(x\right)\right|}^{2}dx=1$

The wave function is defined only in the region $0≤x≤L$therefore substitute csin$\left(\frac{2\mathrm{Ï€°ù}}{L}\right)$for thye second integral and zero for the rest of regions

$$\begin{gathered} 0+{∫}_0^L{\left[c\sin \left(\frac{2\mathrm{Ï€³\ae }}{L}\right)\right]}^{2}dx+0=1 \\ {c}^2{∫}_0^L{\left[c\sin \left(\frac{2\mathrm{Ï€³\ae }}{L}\right)\right]}^{2}dx=1 \end{gathered}$$

consider the following trigonometric relation

${\sin }^2\mathrm{θ}=\frac{1-\cos \mathrm{θ}}{2}$

Hence substitute $\frac{1-\cos \left[2\left({\displaystyle \frac{2\mathrm{Ï€³\ae }}{L}}\right)\right]}{2}$ for sin²$\left(\frac{2\mathrm{Ï€³\ae }}{L}\right)$AND SOLVE FOR C

$$\begin{gathered} C^2{∫}_0^L\frac{1-\cos \left[2\left({\displaystyle \frac{2\mathrm{Ï€³\ae }}{L}}\right)\right]}{2}dx=1 \\ {C}^2{∫}_0^{L}1-\cos \left(\frac{4\mathrm{Ï€³\ae }}{L}\right)dx=1 \end{gathered}$$

$$\begin{gathered} {\left[x\right]}_0^L-{\left[\frac{\sin \left({\displaystyle \frac{4\mathrm{Ï€³\ae }}{L}}\right)}{{\displaystyle \frac{4\mathrm{Ï€}}{L}}}\right]}_0^L=\frac{2}{C^2} \\ L-\frac{L}{4\mathrm{Ï€}}\left[\sin \left(\frac{4\mathrm{Ï€³\ae }}{L}\right)-\sin 0\right]=\frac{2}{C^2} \\ L-0=\frac{2}{C^2} \\ C=\sqrt{\frac{2}{L}} \end{gathered}$$

$$\begin{gathered} P=∫{\left|\mathrm{ψ}\left(x\right)\right|}^{2}dx \\ P={∫}_0^{\frac{L}{3}}{\left[\sqrt{\frac{2}{L}\sin \left(\frac{2\mathrm{Ï€³\ae }}{l}\right)}\right]}^{2}dx \\ =\frac{1}{L}\left[\frac{L}{3}-\frac{L}{4\mathrm{Ï€}}\left(-0.866\right)\right] \\ =0.402 \\ =40.2\% \end{gathered}$$

THEREFORE, THE VALUE OF PROBABILITY IS 40.2%