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In probability theory, a probability density function (PDF) describes the likelihood of a random variable taking on a particular value. In our scenario, the random variable is the distance, or radius, from the center of a circular target to the point of impact. The PDF is crucial in defining how probabilities are distributed within the range of possible values of the distance.
The challenge here is to express this PDF in such a way that it reflects the distribution of hits on the circular target. Since each part of the target is equally likely to be hit by the bullet, the hits create a uniform distribution over the target's area.
This uniform distribution implies that the density of hits per unit area of the circle is constant. Consequently, our probability density function for the radius can be expressed as a linear function: \(D(r) = kr\), where \(k\) is a constant that needs to be determined. This expression indicates that the probability of hitting a point is proportional to the distance \(r\) itself.

The term 'uniform distribution' signifies that every part of the target's surface area is equally likely to be hit. For a circular target, this makes sense because the person firing the gun has no directional bias – being blindfolded ensures there's no preference for any particular direction.
Uniform distribution over the target means that the density per unit area does not change across the surface of the circle. However, when considering the probability density concerning radius, the relationship is slightly different. The area of a circle segment, which is a function of the radius squared, grows with the square of the distance from the center: \( \pi r^2 \).
Thus, while the distribution is uniform over the area, when considering radial distance, it leads to a linear increase in probability density function as expressed in \(D(r) = kr\), formulating the distance-related PDF with a linear increase as the radius increases.

A fundamental principle of probability is that the sum of all probabilities in a distribution should equal 1. In our case, this means that the total probability of hitting the target across its entire surface must be 1.
To ensure this, we apply the normalization condition to the probability density function \(D(r) = kr\). This involves integrating \(D(r)\) over all possible values of the radial distance \(r\), from 0 to \(b\), the radius of the circle:
\[ \int_0^b D(r) \, dr = 1 \]
Substituting \(D(r) = kr\), and solving the integral:
\[ \int_0^b kr \, dr = \left[ \frac{kr^2}{2} \right]_0^b = \frac{kb^2}{2} = 1 \]
This integration determines the value of \(k\), giving us \(k = \frac{2}{b^2}\) by solving \(\frac{kb^2}{2} = 1\). With this value of \(k\), the PDF is now properly normalized, meaning the total probability of hitting the target is equal to 1.

Calculating the average or expected value of a random variable provides insight into its central tendency. In our exercise, we want to find the average distance from the center of the circle to the point where the bullet hits.
The average value of the distance \(r\) can be found using the probability density function \(D(r)\). The formula for the average value is:
\[ \int_0^b rD(r) \, dr \]
Substituting our expression for the density function \(D(r) = \frac{2r}{b^2}\), this becomes:
\[ \int_0^b r \left( \frac{2r}{b^2} \right) \, dr = \frac{2}{b^2} \int_0^b r^2 \, dr \]
This integration results in:
\[ \frac{2}{b^2} \left[ \frac{r^3}{3} \right]_0^b = \frac{2b^3}{3b^2} = \frac{2b}{3} \]
The calculation reveals that the average distance at which the target is hit is \(\frac{2b}{3}\). This average is greater than \(\frac{b}{2}\), indicating that because more area is available further from the center, hits are more likely to occur at larger distances.