Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 8 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

The two outcomes are not independent. Conditional on the selected die, the rolls are independent, but sharing the unknown die creates statistical dependence. Each nonfavored face has probability

(1 - 1/2) /5 = 1/10

.

The unconditional marginal distribution assigns the following probability to a

1

on either roll:

P (X = 1) = P (Y = 1) = \frac{1}{6} (\frac{1}{2} + 5 \cdot \frac{1}{10}) = \frac{1}{6} .

However, by the law of total probability,

P (X = 1, Y = 1) = \frac{1}{6} [(\frac{1}{2})^{2} + 5 (\frac{1}{10})^{2}] = \frac{1}{20} \neq = \frac{1}{36} .

Thus the joint probability does not factor into the two marginals. Conditional independence given a shared latent choice does not imply unconditional independence.

ii

Solution

Let

D

denote the selected die. Bayes' theorem gives

P (D = 1 ∣ X = 1) = \frac{P ( X = 1 ∣ D = 1 ) P ( D = 1 )}{P ( X = 1 )} = \frac{( 1/2 ) ( 1/6 )}{1/6} = \frac{1}{2} .

Only the first roll is conditioned on here; the second outcome has not been observed.

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