Answer to Assignment 2

Answer to Assignment 2:

1. Here’s what we are given:

P(disease) = 1/400 = 0.0025

P(no disease) = 399/400 = 0.9975

P( + | no disease) = 0.02 (these are false positives)

P( – | disease) = 0.03 (these are false negatives)

P( + | disease) = 0.97 (these are true positives)

So we can apply Bayes’ Rule as follows:

Or in words, the probability that a randomly selected person who got a positive test result actually has the disease is only about 10.8%.

2. Here’s what we are given:

P(#1 has gold) = 1/3

P(#1 has goat) = 2/3

P(host shows #2 has goat | #1 has gold) = 1/2 (true regardless of Monty’s rule)

P(host shows 2 has goat | #1 goat) = 1/4 (true if Monty always chooses which other door to open by flipping a coin; this is true because door #2 has a 1/2 chance of having the second goat and a 1/2 chance of being opened, so multiply these to get 1/4)

So we can apply Bayes’ Rule as follows:

Numerator = P(host shows #2 has goat | #1 has gold)∙P(#1 has gold) = (1/2)(1/3) = 1/6

Denominator = P(host shows #2 has goat | #1 has gold)∙P(#1 has gold)

+ P(host shows #2 has goat | #1 has goat)∙P(#1 has goat)

= (1/2)(1/3) + (1/4)(2/3) = 1/3

P(#1 has gold | host shows #2 has goat) = (1/6) / (1/3) = 1/2

Thus, the chance the gold is behind door #1 is 1/2, and the chance it’s behind the other remaining door (#3) is also 1/2. Your odds of winning remain the same whether you switch or not. Importantly, this is true only if Monty follows the random choice rule, meaning there’s a chance he could have opened a door with gold. In the original description of the problem, he would only open a door with a goat.