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Question 1:

There is a fair coin (one side heads, one side tails) and an unfair coin (both sides tails). You pick one at random, flip it 5 times, and observe that it comes up as tails all five times. What is the chance that you are flipping the unfair coin?

Notations: B: baised_coin
F: fair_coin
H: heads
T: tails
5T: five tails

Given:
     P(B) = P(F) = 1/2

     Because P(T / F) = 1/2 = 0.5
     ,therefore P(5T / F) = (1 / 2)^5 = 0.03125

     Because P(T / B) = 1
     ,therefore P(5T / B) = 1

Objective:

P(B / 5T) = ?

As per the Baye's rule:
P(B / 5T) = [P(5T / B) * P(B)] / [P(5T / B) * P(B) + P(5T / F) * P(F)]
i.e. P(B / 5T) = [1 * 0.5] / [1 * 0.5 + 0.03125 * 0.5]
P(B / 5T) = 0.5 / 0.515625
P(B / 5T) = 0.9696969696969697
P(B / 5T) ~ 0.97

Therefore, there is 97 % that the coin is baised one.

Question 2:

Say we have X ~ Uniform(0, 1) and Y ~ Uniform(0, 1). What is the expected value min(X, Y) ?

Answer:

Let take 
    Z = min(X, Y)

    if we take any sample of Z i.e z as the upper bound to fullfill the condition
    then CDF = P(Z <=z )  = P(X <= z, Y <=z)
                          = 1 - P(X > z, Y > z)
    if X and Y are independent events then 
        P(X > z, Y > z) = P(X > z) * P(Y > z)

        therefore, 
        P(Z <=z ) = 1 - P(X > z, Y > z)
                  = 1 - [(1 - z) * (1 - z)]
                  = 1 - (1- z) ^ 2

    Thefore, 
    PDF(z)  = CFD(z) / dz
            = 2 * (1 - z)
    
    and the first expected value of z

    E(z) = Integral(z * PDF(z)) from range 0, 1 = [(1/z^2) - (1-z^3)] from range 0, 1
    E(z)     = 1/3