nruns=1000000
B=runif(nruns,0,1)
C=runif(nruns,0,1)


#P(B+C) < 1/2

s = B + C

length(s[s<.5])/length(s) #.125
## [1] 0.124996
#P(BC) < 1/2

Z = B * C

length(Z[Z<.5])/length(Z)  #>.85
## [1] 0.846843
#Max and Min(B,C)

Q=rep(0, nruns)
R=rep(0, nruns)

for (i in 1: nruns){ 
  Q[i]=max(B[i],C[i]) 
  R[i]=min(B[i],C[i]) 
  }

length(Q[Q<.5])/length(Q) #.25
## [1] 0.250228
length(R[R<.5])/length(R) #.75
## [1] 0.750574
#visuals
par(mfrow=c(3,2))
hist(B+C)
hist(B*C)
hist(abs(B-C))
hist(Q)
hist(R)

a) B + C < 1/2

From the given information, \(B ~ U[0,1]\) and \(C ~ U[0,1]\).

The density of \(B\) is \(f(b) = 1\), for \(0 \leq b \leq 1\)

The distribution function of \(B\) is \[F(b) = P(B \leq b)\] \[\int_{0}^{b}f(b)db\] \[\int_{0}^{b}1db\] \[=b\] The density of \(C\) is \(f(c) = 1\), for \(0 \leq c \leq 1\)

The distribution function of \(C\) is \[F(c) = P(C \leq b)\] \[\int_{0}^{c}f(c)dc\] \[\int_{0}^{c}1dc\] \[=c\] Since B and C are independent, the joint density of B & C is \[f(b,c) = f(b) * f(c)\] \[1*1\] \[=1\]

\[P\bigg(B+C<\frac{1}{2}\bigg)\] \[P\bigg(X+Y<\frac{1}{2}\bigg)\] \[P\bigg(0<X<\frac{1}{2}, 0<Y<\frac{1}{2}-x\bigg)\] \[\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}-x} f(x,y)dxdy\] \[\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}-x} 1dydx\] \[\int_{0}^{\frac{1}{2}} \bigg[y\bigg]_{0}^{\frac{1}{2}-x}dx\] \[\int_{0}^{\frac{1}{2}} \bigg[\frac{1}{2}-x\bigg]dx\] \[\bigg[\frac{1}{2}x-\frac{x^2}{2}\bigg]_{0}^\frac{1}{2}\] \[\frac{1}{4}-\frac{1}{8}\] \[=\frac{1}{8}\] \[=0.125\]

b) BC < 1/2

Let \(Z=BC\) or \(Z = XY\) \[P\bigg(BC < \frac{1}{2}\bigg) = P\bigg(Z < \frac{1}{2}\bigg)\] \[\int_0^\frac{1}{2} f(z) dz\] \[\int_0^\frac{1}{2} -ln(z)dz\] \[-\bigg[z ln(z)-z\bigg]_0^\frac{1}{2}\] \[=0.8466\]

c) |B ??? C| < 1/2

\(P\bigg(\lvert B-C \rvert < \frac{1}{2}\) given that \(0<x+y<1\bigg)\) \(P\bigg(\lvert x-y \rvert < \frac{1}{2}\) given that \(0<x+y<1\bigg)\) \[P\bigg(-\frac{1}{2} < x-y < \frac{1}{2} ; 0 < x < 1-y\bigg)\] \[P\bigg(y-\frac{1}{2} < x < y+\frac{1}{2} ; 0 < x < 1-y\bigg)\] \[P\bigg(0 < x < \frac{1}{2} - y, 0 < y < \frac{1}{2}\bigg)\]

\[\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}-y} f(x,y)dxdy\]

\[\int_{0}^{\frac{1}{2}} \int_{0}^{\frac{1}{2}-y} 1dxdy\]

\[\int_{0}^{\frac{1}{2}} \bigg[x\bigg]_{0}^{\frac{1}{2}-y}dy\] \[\int_{0}^{\frac{1}{2}} \bigg[\frac{1}{2}-y\bigg]dy\] \[\bigg[\frac{y}{2}-\frac{y^2}{2}\bigg]_{0}^\frac{1}{2}\]

\[\frac{\frac{1}{2}}{2}-\frac{\frac{1}{2}}{2}^2\] \[\frac{1}{4} - \frac{1}{8}\] \[=\frac{1}{8}\] \[=0.125\]

d) max{B,C} < 1/2

Let max\({B,C} < \frac{1}{2}\) be an event, then
\[P\bigg( max(B,C) < \frac{1}{2} \bigg) = P\bigg(B < \frac{1}{2}, C<\frac{1}{2}\bigg)\] \[P\bigg(B<\frac{1}{2}\bigg)P\bigg(C<\frac{1}{2}\bigg)\] \[\frac{1}{2} * \frac{1}{2}\] \[=\frac{1}{4}\] \[=0.25\]

e) min{B,C} < 1/2

Let min\({B,C} < \frac{1}{2}\) be an event, then
\[P\bigg( min(B,C) < \frac{1}{2} \bigg) = 1-P\bigg(min(B,C) > \frac{1}{2}\bigg)\] \[1-P\bigg(B > \frac{1}{2}, C>\frac{1}{2}\bigg)\] \[1-P\bigg(B > \frac{1}{2}\bigg) P\bigg( C>\frac{1}{2}\bigg)\] \[1-\bigg( 1-P\bigg(B \leq \frac{1}{2} \bigg)\bigg)\bigg(1-P\bigg(C \leq \frac{1}{2} \bigg) \bigg)\] \[1-\bigg(1-\frac{1}{2}\bigg)\bigg(1-\frac{1}{2}\bigg)\] \[1-\bigg(1-\frac{1}{2}\bigg)^2\] \[1-\frac{1}{4}\] \[=\frac{3}{4}\] \[=0.75\]