Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 8

The single-roll colour probabilities are $1/2$, $1/3$ and $1/6$. There are $3!$ possible orders of the three distinct colours, and the independent rolls give each order probability $(1/2)(1/3)(1/6)$. The events are disjoint, so the required probability is

The all-red event $B$ is a subset of the same-colour event $A$. Its probability is $(1/2{)}^3=1/8$, while the three disjoint monochromatic cases giveThus the conditional probability isThe monochromatic outcomes are not equally likely, so counting colours alone would give the wrong conditional answer.

Let $I_c$ be the indicator random variable that colour $c$ appears at least once. Then $X=I_R+I_B+I_G$. For a colour of single-roll probability $p_c$, independence of rolls gives $\mathbb{P}(I_c=0)=(1-p_c{)}^n$. By linearity of expectation,The three indicators are not independent, but linearity of expectation does not require them to be. This is the expected number of observed categories.

For $x>1$, write the proposed density as $x^{-\alpha }(c-1/x)$. Since the first factor is positive, nonnegativity for all such $x$ requires $c\ge 1$; if $c<1$, values sufficiently close to one make it negative. With $c\ge 1$, the positive tail is integrable only if $\alpha >1$. In that case normalization requiresImposing $c\ge 1$ now gives ${\alpha }^2-\alpha -1\ge 0$. Combining this with $\alpha >1$ yieldsConversely, these conditions ensure a nonnegative, integrable function with integral one, so they are sufficient as well as necessary. The endpoint is allowed: the density vanishes at the left-hand boundary limit, not on its whole support.

Normalization gives $c=8/3$. Integrating the probability density from one to two givesThe expected value exists, since the first-moment integrals converge, and equals