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You are extremely close to the answer. You have already noted that the area of a circle with radius $r$ is $A(r) = \pi r^2$. Since you are selecting a uniform point on the circle, this means that you have the cumulative distribution function:

$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2.$$

The reasoning behind this is that with a uniform distribution we have a flat density over the circle. So the probability of falling within a given area is the relative size of that area compared to the size of the circle. This gives the corresponding density function:

$$f_R(r) = 2r \quad \quad \text{for all } 0 \leqslant r \leqslant 1.$$

As you can see from this result, although the chosen point is uniformly distributed on the circle, the resulting distance $R$ to the point does not have a uniform distribution; as expected, it is more likely to be nearer to one than nearer to the outer edge of the circle than its origin.

You are extremely close to the answer. You have already noted that the area of a circle with radius $r$ is $A(r) = \pi r^2$. Since you are selecting a uniform point on the circle, this means that you have the cumulative distribution function:

$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2 \quad \quad \quad 0 \leqslant r \leqslant 1.$$

The reasoning behind this is that with a uniform distribution we have a flat density over the circle. So the probability of falling within a given area is the relative size of that area compared to the size of the circle. This gives the corresponding density function:

$$f_R(r) = 2r \quad \quad \text{for all } 0 \leqslant r \leqslant 1.$$

As you can see from this result, although the chosen point is uniformly distributed on the circle, the resulting distance $R$ from the origin to the point does not have a uniform distribution; as expected, it is more likely to be nearer to the outer edge of the circle than its origin.

You are extremely close to the answer. You have already noted that the area of a circle with radius $r$ is $A(r) = \pi r^2$. Since you are selecting a uniform point on the circle, this means that you have the cumulative distribution function:

$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2.$$

The reasoning behind this is that with a uniform distribution we have a flat density over the circle. So the probability of falling within a given area is the relative size of that area compared to the size of the circle. This gives the corresponding density function:

$$f_R(r) = 2r \quad \quad \text{for all } 0 \leqslant r \leqslant 1.$$

As you can see from this result, although the chosen point is uniformly distributed on the circle, the resulting distance $R$ to the point does not have a uniform distribution; as expected, it is more likely to bear nearer to one than nearer to zero.

You are extremely close to the answer. You have already noted that the area of a circle with radius $r$ is $A(r) = \pi r^2$. Since you are selecting a uniform point on the circle, this means that you have the cumulative distribution function:

$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2.$$

The reasoning behind this is that with a uniform distribution we have a flat density over the circle. So the probability of falling within a given area is the relative size of that area compared to the size of the circle. This gives the corresponding density function:

$$f_R(r) = 2r \quad \quad \text{for all } 0 \leqslant r \leqslant 1.$$

As you can see from this result, although the chosen point is uniformly distributed on the circle, the resulting distance $R$ to the point does not have a uniform distribution; as expected, it is more likely to be nearer to one than nearer to the outer edge of the circle than its origin.

You are extremely close to the answer. You have already noted that the area of a circle with radius $r$ is $A(r) = \pi r^2$. Since you are selecting a uniform point on the circle, this means that you have the cumulative distribution function:

$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2.$$

This gives the corresponding density function:

$$f_R(r) = 2r \quad \quad \text{for all } 0 \leqslant r \leqslant 1.$$

You are extremely close to the answer. You have already noted that the area of a circle with radius $r$ is $A(r) = \pi r^2$. Since you are selecting a uniform point on the circle, this means that you have the cumulative distribution function:

$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2.$$

The reasoning behind this is that with a uniform distribution we have a flat density over the circle. So the probability of falling within a given area is the relative size of that area compared to the size of the circle. This gives the corresponding density function:

$$f_R(r) = 2r \quad \quad \text{for all } 0 \leqslant r \leqslant 1.$$

As you can see from this result, although the chosen point is uniformly distributed on the circle, the resulting distance $R$ to the point does not have a uniform distribution; as expected, it is more likely to bear nearer to one than nearer to zero.