Quantitative proficiency test

26. total 52 cards, 26 is red, another 26 is black, 2 of special is red, 2 of the special is black, total red + special = red + special(black) = 26 + 2 = 28
    ans : 28/52
27. 20C6 * 30C9 / 52C15
28. integration of p.d.f lead to [-c/x^2] from 25 to infinity, and we can get 1/25
29. integration of x times fx(x) = 6(1/3)x^3 - 6(1/4)x^4 from 0 to 1, so E(x) will be 2 - 1.5 = 0.5
    while for Variance, it equal to E(x^2) - [E(X)]^2, so integration of x^2 times fx(x) = 6/4 x^4 - 6/5 x^5 from 0 to 1, and lead to 6/20.
    Var(x) = 6/20 - (1/2)^2 = 6/20 - 5/20 = 1/20
30. integrate e(-y-x) by using dy from 0 to x and dx from 0 to infinity, the integrated possibility = e(-x^2)/2 from 0 to infinity, = 1/2 - 0 = 1/2
31. integrate 2 by using dx from 0 to y we get fy(y), which equal to 2y. then fxy(x,y)/fy(y) = fx|y(x|y)= 2/2y = 1/y. integrate  fx|y(x|y) by using dx from 0.5 to y,
    we get (y-0.5)/y, then substitute y = 0.7 into the equation, answer will be (0.7-0.5)/0.7 = 2/7.
32. fy(y) = integrate 4xy by dx from 0 to 1 = 4x^2y/2 = 2y, in the same way, fx(x) = 2x, cov(x,y) = E[XY] - E[X]E[Y], = integrate xy(4xy) by using dx and dy from 0 to 1,
    minus the integral of x(2x) and y(2y), = 4/9 - (2/3)*(2/3) = 0
33. P(X=0) = Gx(0) = 1/(2-0) = 1/2, P(X=1) = G'x(0) = 1/(2-0)^2 = 1/4, 1/2+1/4 = 3/4
34. For moment generating function, E[x] = first derivative of the function with regards of t = 0, so E[x] = 1/(1-0)^2 = 1, Var(x) = E[x^2] - (E[x])^2, E[x^2] = second
    derivative of the function = 2/(1-0)^3 = 2, Var(x) = 2 - (1)^2 = 1.