Question

f(x) = 3x, 0<x<1 0, otherwise Problem 9: Find the CDF of X, [4] Problem 10: Find P(X < 1/3) [4] Problem 11: Find the 95th percentile of X (Remark: You have to solve for x in the equation P(X < x) = .95) [4] Problem 12: Find P(X > 2/3|X > 1/3) [4] A pipe-smoking Mathematician carries two match boxes, box A in his left hand and box B in right-hand pocket. Initially, each box contains 3 matches. Each time the Mathematician requires a match, he is 2/3 likely to take it from box A. At the moment he first discovers that box A to be empty. Problem 13: How many matches are expected to be left in the other box? [4]

Answer 1

We would be looking at the first 4 questions here as:

Q9) The CDF for X here is obtained as:

$F_x(x) = P(X \leq x) = \int_{0}^{x}f(t) \ dt = \int_{0}^{x} 3t \ dt = \frac{3x^2}{2}$

Therefore the CDF for X here is given as:

$F_x(x) = \frac{3x^2}{2}, 0 < x < 1$

Q10) The probability here is computed as:

$P(X < 1/3) = F_x(1/3) = \frac{3(1/3)^2}{2} = \frac{1}{6}$

Therefore 1/6 = 0.1667 is the required probability here.

Q11) Let the 95th percentile value of X here be K. Then, we have here:

F(K) = 0.95

$\frac{3K^2}{2} = 0.95$

K = 0.7958

Therefore 0.7958 is the required 95th percentile value here.

Q12) The conditional probability here is computed using Bayes theorem as:

$P(X > 2/3 | X > 1/3) = \frac{P(X > 2/3)}{P(X > 1/3)}$

$P(X > 2/3 | X > 1/3) = \frac{1 - F(2/3)}{1 - F(1/3)}$

$P(X > 2/3 | X > 1/3) = \frac{1 - 1.5*(2/3)^2}{1 - 1.5*(1/3)^2} = 0.4$

Therefore 0.4 is the required probability here.