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Choose independently two numbers B and C at random from the interval [0, 1] with uniform density. Prove that B and C are proper probability distributions. Note that the point (B,C) is then chosen at random in the unit square.

B <- runif(100000, min = 0, max = 1)
max(B)

## [1] 0.9999977

min(B)

## [1] 6.437767e-07

C <- runif(100000, min = 0, max = 1)
max(C)

## [1] 0.9999926

min(C)

## [1] 1.12066e-05

(a) B+C < 1/2

a <- sum((B+C) < 0.5)/100000
a

## [1] 0.12358

(b) BC < 1/2

b <-sum((B*C) < 0.5)/100000
b

## [1] 0.84609

(c) |B-C| < 1/2

c <-sum(abs(B-C) < 0.5)/100000
c

## [1] 0.74829

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

P(B < 1/2) = 0.5 and P(C < 1/2)=0.5, the probability that both are less than 1/2 is 0.5*0.5 = 0.25. So the only way the maximum of B and C are not greater than 0.5 is if both are less than 0.5, which is the compliment of both being less than 0.5, or 1??? 0.25 = 0.75.

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

For the minimum of B or C to be less than 1/2, both B and C must be less than 0.5. Because the probability that P(B < 1/2) = 0.5 and P(C < 1/2) = 0.5, the probability that both are less than 1/2 is 0.5*0.5 = 0.25.