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Solution

1. There are different ways of setting this up, but a straight-forward way is to model the basic outcomes as pairs [x,y], where x,y {H,T} are the outcomes of the first and second flip respectively (H being heads, and T being tails). That is:

   Ω = { [H,H], [H,T], [T,H], [T,T]}

2. Here are the two distributions, which I call p₁ and p₂:

   • p₁([H,H]) = p₁([H,T]) = p₁([T,H]) = p₁([T,T]) = 1/4

   • p₂([H,H]) = 4/9, p₂([H,T]) = 2/9, p₂([T,H]) = 2/9, p₂([T,T]) = 1/9

   For p₁, it should be pretty clear how to arrive at these values: each outcome is equally likely and there are four outcomes, so they must all have probability 1/4.

   For p₂, one way you can arrive at the values is to treat heads and tails as independent events and the pair as their conjunction. That is we assign H = 2/3 and T = 1/3 to satisfy the constraint, and then calculate [H,T] = 2/3 x 1/3 = 2/9, etc.

3. First, let’s translate the events into set representations:

   • “At least one flip is heads” becomes {[H,H], [H,T], [T,H]}

   • “Both flips are tails” becomes {[T,T]}

   • “The first flip is heads or the second is tails” becomes {[H,T], [H,H], [T,T]}

   • “None of the two flips is heads” becomes {[T,T]}.

   For the probabilities of these events under the distributions, we simply add up the weights of the events:

   • Pr₁({[H,H], [H,T], [T,H]}) = p₁([H,H]) + p₁([H,T]) + p₁([T,H]) = 1/4 + 1/4 + 1/4 = 3/4

   • Pr₂({[H,H], [H,T], [T,H]}) = p₂([H,H]) + p₂([H,T]) + p₂([T,H]) = 4/9 + 2/9 + 2/9 = 8/9

   • Pr₁({[T,T]}) = p₁([T,T]) = 1/4

   • Pr₂({[T,T]}) = p₂([T,T]) = 1/9

   • Pr₁({[H,T], [H,H], [T,T]}) = p₁([H,T]) + p₁([H,H]) + p₁([T,T]) = 1/4 + 1/4 + 1/4 = 3/4

   • Pr₂({[H,T], [H,H], [T,T]}) = p₂([H,T]) + p₂([H,H]) + p₂([T,T]) = 2/9 + 4/9 + 1/9 = 7/9

Solution

We start with the trivial identity:

Since A is equivalent to (A B) (A B), we can infer:

Pr(A) = Pr((A B) (A B))

We observe that A B) and A B are logically incompatible, that is (Pr((A B) (A B))). Thus:

Pr((A B) (A B)) = Pr((A B) + (A B))

So, we arrive at:

Pr(A) = Pr((A B) + (A B))

Replacing the conjunctive probabilities using the conjunction rule, we get:

Pr(A) = Pr(A | B)Pr(B) + Pr(A | B)Pr( B)

Solution

Suppose that A B. For A B to be the case, we need that Pr(B | A) ≥ Pr(B). Unfolding the definition of conditional probability, we get:

Pr(B | A) = Pr(B A)/Pr(A)

But since A B, we have that Pr(B A) = Pr(A). But that means that:

But since all probabilities are less than 1, it follows that Pr(B | A) ≥ Pr(A).

Solution

1. Here’s the relevant probabilities:

   • Pr(HasDisease) = 0.01
   • Pr(TestPositive | HasDisease) = 0.9
   • Pr(¬TestPositive | ¬HasDisease) = 0.9

2. According to Bayes formula

   Pr(HasDisease | TestPositive)
   =
   [Pr(HasDisease) x Pr(TestPositive | HasDisease) ] / Pr(TestPositive)

   We have Pr(HasDisease) and Pr(TestPositive | HasDisease), but we need to figure out Pr(TestPositive).

   For this, we use the law of total probability. Apply it to TestPositive as A and HasDisease as B gives us:

   Pr(TestPositive)
   =
   Pr(TestPositive| HasDiesease)Pr(HasDisease)
   +
   Pr(TestPositive | ¬HasDisease)Pr(¬HasDisease)

   We have Pr(TestPositive| HasDiesease) and Pr(HasDisease) given. To obtain Pr(TestPositive | ¬HasDisease) and Pr(¬HasDisease), we apply the negation laws:

   Pr(TestPositive | ¬HasDisease) = 1 - Pr(¬TestPositive | ¬HasDisease)
   Pr(¬HasDisease) = 1 - Pr(HasDisease)

   Now we have all the relevant values and can calculate:

   Pr(¬HasDisease) = 1 - Pr(HasDisease) = 1 - 0.01 = 0.99
   Pr(TestPositive | ¬HasDisease) = 1 - Pr(¬TestPositive | ¬HasDisease)
   = 1 - 0.9 = 0.1
   Pr(TestPositive | ¬HasDisease)Pr(¬HasDisease) = 0.99 x 0.1 = 0.099

   So:

   Pr(TestPositive)
   =
   Pr(TestPositive| HasDiesease)Pr(HasDisease)
   +
   Pr(TestPositive | ¬HasDisease)Pr(¬HasDisease)
   =
   (0.9 x 0.01) + 0.099 = 0.108

   Now for the final probabilities:

   Pr(HasDisease | TestPositive)
   =
   [Pr(HasDisease) x Pr(TestPositive | HasDisease) ] / Pr(TestPositive)
   =
   (0.01 x 0.9) / 0.108 = 0.009 / 0.108 = 0.083...

   In other words, a random person testing positive, given this setup, gives us a probability of around 8% of them having the disease—even if the test is 90% reliable. This is because the prior probability of the person having the disease—the so-called base-rate—is very low.

   Note that it’s important here that the person was chosen randomly. If there is additional information, such as them showing symptoms, the prior of them having the disease would be different.

3. If we re-set the probabilities using Bayesian updating, we now have:

   Pr(Disease) = 0.083...

   This affects the value of Pr(TestPositive), which we’ve calculated using Pr(Disease).

   We now get:

   Pr(¬HasDisease) = 1 - Pr(HasDisease) = 1 - 0.083.. = 0.916...
   Pr(TestPositive | ¬HasDisease)Pr(¬HasDisease) = 0.1 x 0.916... = 0.0916...

   Similarly, we get for the other summand:

   Pr(TestPositive | HasDisease)Pr(HasDisease) = 0.9 x 0.083.. = 0.0749...

   So, we now have for Pr(TestPositive) that:

   Pr(TestPositive) = 0.0749... + 0.0916... = 0.16...
   Pr(HasDisease | TestPositive)
   =
   [Pr(HasDisease) x Pr(TestPositive | HasDisease) ] / Pr(TestPositive)
   =
   (0.083 x 0.9) / 0.16... = 0.0747 / 0.16... ≈ 0.47

   So, we need to do another test. Now we go somewhat quicker:

   Pr(Disease) = 0.47
   Pr(TestPositive | ¬HasDisease)Pr(¬HasDisease) = 0.1 x 0.53 = 0.053
   Pr(TestPositive | HasDisease)Pr(HasDisease) = 0.9 x 0.46 = 0.414

   So:

   Pr(TestPositive) = 0.467
   Pr(HasDisease | TestPositive)
   =
   [Pr(HasDisease) x Pr(TestPositive | HasDisease) ] / Pr(TestPositive)
   =
   (0.47 x 0.9) / 0.467... ≈ 0.9

   So only after the third positive test in a row, we can infer with 90% confidence that the patient has a disease.