Probability 難題 (2)

F(x) = integrate 6x(1-x) dx
= 6 integrate(x - x^2) dx
= 6 (x^2/2 - x^3/3)
= 3x^2 - 2x^3

P(X < 1/4) = F(1/4)
= 3(1/4)^2 - 2(1/4)^3
= 5 / 32

P(X > 1/2) = 1 - P(X < 1/2) = 1 - F(1/2)
= 1 - (3(1/2)^2 - 2(1/2)^3)
= 1/2

For this part, one may argue that 0.5 is actually the axis of symmetry and avoid the integration.

2a) There are four cases:

XX (with prob = 48/52 * 47/51 = ), at that time, P(Z = 0, W = 0) = 188/221
XA (with prob = 48/52 * 4/51), at that time, P(Z = 0, W = 1) = 16/221
AX (with prob = 4/52 * 48/51), at that time, P(Z = 1, W = 1) = 16/221
AA (with prob = 4/52 * 3/51), at that time, P(Z = 1, W = 2) = 1/221

[Where X denotes non Ace]

That is the joint prob of Z and W, all other cases has zero probability.

2b) P(Z = 0) = 48/52 = 12/13, P(Z = 1) = 4/52 = 1/13, that can be obtained inituitively, or you can marginalize the joint probabilities.

P(Z = 0) = P(Z = 0, W = 0) + P(Z = 0, W = 1) + P(Z = 0, W = 2) = 188/221 + 16/221 + 0 = 12/13
P(Z = 1) = P(Z = 1, W = 0) + P(Z = 1, W = 1) + P(Z = 1, W = 2) = 0 + 16/221 + 1/221 = 1/13

2c) P(W = 0| Z = 1) = 0, P(W = 1| Z = 1) = 48/51 = 16/17, P(W = 2|Z = 1) = 3/51 = 1/17. This can be obtained inituitively, or you can use the definition on conditional probability

P(A = x| B = y) = P(A = x, B = y) | P (B = y), e.g.

P(W = 1| Z = 1) = P(W = 1, Z = 1) / P(Z = 1) = (16/221)/(1/13) = 16/17.