B and C are proper probability distributions.

It suffices to prove the \(B\) has a proper probability distribution since \(B\) and \(C\) are identically distributed. We will show the uniform density on the interval \([0,1]\) is a probability distribution.

\[f_B(x) = 1 \text{ for } x \in [0,1]\]

Clearly \[ f_B(x) \geq 0\]. and \[\int_0^1 f_B(x)dx = x |_{x=0}^{x=1} = 1 - 0 = 1\]

Part a. \(B + C < 1/2\)

The probability of \(B+C<1/2\) is equal to the area of the region bounded by 3 lines: \(B=0\) the vertical axis, \(C=0\) the horizontal axis and the diagonal line \(B + C = 1/2\). This forms the right triangle whose vertices are \((0,1/2), (1/2,0), (0,0)\).

Shaded triangle satisfies the inequality:

This has an area of \[\frac{1}{2}ab = \frac{1}{8} = Prob( B + C < \frac{1}{2})\]

Part b. \(BC < 1/2\)

The probability of \(BC<1/2\) is equal to the area bounded by 5 lines: \(B=0\), the vertical axis, \(C=0\) the horizontal axis and vertical line \(B=1\) and the curve \(BC=1/2\) and the horizontal line \(C=1\). This forms a 5 sided region bounded by the points in order counter clock-wise: \[(0,0), (1,0), (1 , \frac{1}{2}), (\frac{1}{2}, 1), (0,1)\].

The area of this region can be solved by finding areas of two subregions. The first subregion is the area from \(B=0\) to \(B=1/2\) which is \(\frac{1}{2}\) since this is just a rectange of height 1 and width \(1/2\).

The second subregion requires integration because the upper boundary is a function. \[C = f(B) = \frac{1}{2B}\] between the values \(B=1/2\) to \(B=1\). The area of this second region is:

\[\int_{1/2}^{1} f(B)dB = \int_{1/2}^{1} \frac{dB}{2B} = \frac{1}{2}\left[ log(1) - log(\frac{1}{2}) \right] = \frac{1}{2}log(2) \approx 0.3465736\]

Adding this up gives the answer:

\[P( BC < \frac{1}{2}) = \frac{1}{2} + \frac{1}{2}log(2) = \frac{1}{2}(1+log(2)) \approx 0.8465736\].

Part c. \(| B- C | < 1/2\)

The region \(\lvert{ B - C }\rvert < 1/2\) is bound by the boundary of the unit cube and the lines:

\[ B - C = \frac{1}{2} \text{ and } C - B = \frac{1}{2}\] The region formed as shape with vertices in counter-clockwise order:

\[ (0,0), (\frac{1}{2}, 0 ), ( 1, \frac{1}{2} ), ( 1, 1), (\frac{1}{2}, 1), ( 0, \frac{1}{2})\] The probability is the complement of the two right triangles inside the unit square. Each triangle has sides of length 0.5. Thus, each triangle has area \[\frac{1}{8}\]. The two triangles have a sum of areas equal to \(\frac{1}{4}\).

Therefore the probability \[P( \lvert B - C \rvert < 1/2 ) = 1 - \frac{1}{4} = \frac{3}{4}\].

Part d. \(max(B,C) < 1/2\)

The region \(max(B,C) < 1/2\) is the complement inside the unit square of the region \(B>1/2 \text{ and } C> 1/2\). The latter region is a square of side length \(\frac{1}{2}\) are area \(\frac{1}{4}\).

Therefore, the complement \[P( max(B,C) < \frac{1}{2} ) = 1 - \frac{1}{4} = \frac{3}{4}\].

Part e. \(min(B, C) < 1/2\)

The inequality \[min(B, C) < \frac{1}{2} \] is equivalent to \[Pr(min(B,C)<1/2) = Pr\{(B,C) | ( B < 1/2) \text{ or } ( C < 1/2) \} = Pr\{ (B, C) | B > 1/2 \text{ and } C > 1/2 \}^c=1-\frac{1}{4}=\frac{3}{4}\] by a straightforward application of DeMorgan’s Law.

The region is equivalent to the complement of the unit square with the upper right sub-square of length 1/2 removed.