Homework 05

Week05 Problem Set -1
Solution

B and C are independent random variables from the unit interval [0, 1] with the uniform density.
Let \(x\) and \(y\) represent the independent random variables \(B\) and \(C\) respectively from the unit interval \([0, 1]\) with the uniform density.

(a) Find the probability that B+C < 1/2.
Question_5.1_Diagram

Let’s represent \(y\) in terms of \(x\).
\(x+y<\frac{1}{2}\)
\(y=\frac{1}{2}-x\) in the interval \([0, 1]\)
The graph of this function is a triangle as shown above. In order to find the probability, we need to get the area of the shaded region. The base of the triangle is \(\frac{1}{2}\) and the height of the triangle is \(\frac{1}{2}\):

Area of shaded region = \(\frac{1}{2}\) X base X height = \(\frac{1}{2}\)X\(\frac{1}{2}\)X\(\frac{1}{2}\) = \(\frac{1}{8}\)
Therefore, \(P\bigg(B+C < 1/2\bigg) = \frac{1}{8}\)

(b) Find the probability that BC < 1/2.
Question_5.2_Diagram

Let’s represent \(y\) in terms of \(x\).
\(xy<\frac{1}{2}\)
\(y<\frac{1}{2x}\) in the interval \([0, 1]\)

The graph of this function is as shown above. In order to find the probability, we need to get the area of the shaded region. This is a combination of the rectangle \(ABFG\) and the area of region \(BCDFIB\) under the curve.
Area of the shaded region = Area of rectangle \(ABFG\) + Area of region \(BCDFIB\) under the curve
Area of the shaded region = (\(\frac{1}{2}\)X\(1\)) + (\(\int_{0.5}^{1}\frac{1}{2x}dx\)) = \(0.5 + 0.3465 = 0.8465\)

Therefore, \(P\bigg(BC < 1/2\bigg) = 0.8465\)

(c) Find the probability that |B-C| < 1/2.
Question_5.3_Diagram

\(\begin{aligned} \left|x-y\right|<\frac{1}{2} \end{aligned}\) in the interval \([0, 1]\)

The graph of this function is as shown above. In order to find the probability, we need to get the area of the shaded region. This is a combination of the 2 rectangles (\(ABGF\) and \(GCDE\)) and 2 triangles (\(FGE\) and \(BGC\))

Area of the shaded region = Area of rectangle \(ABGF\) + Area of rectangle \(GCDE\) + Area of triangle \(GFE\) + Area of triangle \(BGC\)
Area of the shaded region = (\(\frac{1}{2}\) X \(\frac{1}{2}\)) + (\(\frac{1}{2}\) X \(\frac{1}{2}\)) + (\(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\)) + (\(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2})\) = \(\frac{3}{4}\)

Therefore, \(\begin{aligned}P\bigg(\left|B-C\right|<\frac{1}{2} \bigg) = \frac{3}{4}\end{aligned}\)

(d) Find the probability that max{B,C} < 1/2.
Question_5.4_Diagram

\(\begin{aligned} max(x,y)<\frac{1}{2} \end{aligned}\) in the interval \([0, 1]\)

The graph of this function is as shown above. In order to find the probability, we need to get the area of the shaded region. This is the area of the rectangles \(ABCD\).

Area of the shaded region = Area of rectangle \(ABCD\)
Area of the shaded region = \(\frac{1}{2}\) X \(\frac{1}{2}\) = \(\frac{1}{4}\)

Therefore, \(\begin{aligned}P\bigg(max\{B, C\}<\frac{1}{2} \bigg) = \frac{1}{4}\end{aligned}\)

(e) Find the probability that min{B,C} < 1/2.
Question_5.5_Diagram

\(\begin{aligned} min(x,y)<\frac{1}{2} \end{aligned}\) in the interval \([0, 1]\)

The graph of this function is as shown above. In order to find the probability, we need to get the area of the shaded region. This is the combination of the area of the rectangles \(ABCG\) and the area of the square \(GDEF\).

Area of the shaded region = Area of rectangle \(ABCG\) + Area of the square \(GDEF\)
Area of the shaded region = (\(\frac{1}{2}\) X \(1\)) + (\(\frac{1}{2}\) X \(\frac{1}{2}\)) = (\(\frac{1}{2}\)) + (\(\frac{1}{4}\)) = \(\frac{3}{4}\)

Therefore, \(\begin{aligned}P\bigg(min\{B, C\}<\frac{1}{2} \bigg) = \frac{3}{4}\end{aligned}\)