Question

2.

a) Suppose that a bag contains six slips of paper: one with the number 1 written on it, two with the number 2, and three with the number 3. What is the expected value of the number drawn if one slip is selected at random from the bag? Type your answer as a fraction. Ex: 5/2

b) For the question above, what is the variance of the number drawn if one slip is selected at random from the bag? Type your answer as a fraction. Example: 5/2.

3. What is the probability that a random person who tests positive for a certain blood disease actually has the disease, if we know that 1% of the population has the disease, that 95% of those who have the disease test positive for it, and 2% of those who do not have the disease test positive for it. Type your answer as a decimal to the nearest thousandth. Example 0.233.

Answer 1

a)

Probability of selecting any slip with number 1 , P(X = 1) = 1/6

Probability of selecting any slip with number 2 , P(X = 2) = 2/6

Probability of selecting any slip with number 3 , P(X = 3) = 3/6

Expected value of the number drawn, E(X) = = 1 * (1/6) + 2 * (2/6) + 3 * (3/6) = (1 + 4 + 9)/6 = 14/6 = 7/3

b)

= = (1 + 8 + 27)/6 = 36/6 = 6

variance of the number drawn =

3.

P(disease) = 0.01

P(positive | disease) = 0.95

P(positive | ~disease) = 0.02

By law of total probability,

P(positive) = P(disease) P(positive | disease) + P(~disease) P(positive | ~disease)

= 0.01 * 0.95 + (1 - 0.01) * 0.02

= 0.0293

probability that a random person who tests positive for a certain blood disease actually has the disease

= P(disease | positive) = P(positive | disease)  P(disease) / P(positive) (Bayes Theorem)

= 0.95 * 0.01 / 0.0293

= 0.3242321