Q 5.30 - A First Course in Probability

Question

An image is partitioned into two regions, one white and the other black. A reading taken from a randomly chosen point in the white section will be normally distributed with $\mu =4$and${\sigma }^2=4$, whereas one taken from a randomly chosen point in the black region will have a normally distributed reading with parameters$(6,9)$. A point is randomly chosen on the image and has a reading of$5$. If the fraction of the image that is black is$\alpha$, for what value of$\alpha$ would the probability of making an error be the same, regardless of whether one concluded that the point was in the black region or in the white region? 

Step-by-Step Solution

Answer

Therefore, the value of$\alpha =0.3827$$\alpha =0.3827$.

Step 1 Given information:

A reading taken from a randomly chosen point in the white section will be normally distributed with $\mu =4$and${\sigma }^2=4$, whereas one taken from a randomly chosen point in the black region will have a normally distributed reading with parameters $(6,9)$

Step 2 Explanation:

$P(given point belong to black region)=\frac{1}{2}$

Generally,

Equation 1:

$P(given point belong to black region)=\frac{\alpha \times P(value={\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$black$}\right.})}{\alpha \times P(value={\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$black$}\right.})+(1-\alpha )\times P(value={\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$white$}\right.})}$

Step 3 Explanation:

Let's put them equal to each other:

Put equation 1$$\begin{gathered} \frac{\alpha \times {\displaystyle \frac{e^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}}{2\sqrt{(2\times \mathrm{\pi })}}}}{\alpha \times {\displaystyle \frac{e^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}}{2\sqrt{(2\times \mathrm{\pi })}}}+(1-\alpha ){\displaystyle \frac{e^{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}}{3\sqrt{(2\times \mathrm{\pi })}}}}=\frac{1}{2} \\ \frac{{\displaystyle \frac{\alpha }{2}}}{{\displaystyle \frac{\alpha }{2}}+{\displaystyle \frac{(1-\alpha )\times e^{{\displaystyle \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$72$}\right.}}}{3}}}=\frac{1}{2} \\ \frac{3\alpha }{3\alpha +2(1-\alpha )\times e^{{\displaystyle \raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$72$}\right.}}}=\frac{1}{2} \\ 6\alpha =3\alpha +2(1-\alpha )\times e^{\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$72$}\right.} \\ 3\alpha =1.86-1.86\alpha \\ \alpha =0.3827 \end{gathered}$$