Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 16

Random sampling without replacement assigns each of the ten distinguishable pieces to each lunch with the same probability. Five pieces of each type therefore giveOne can check the second probability by conditioning on the first lunch:The identical marginal probabilities do not imply independence of the draws.

Condition on an apple occupying position two in a uniformly random ordering. The nine other positions then contain four apples and five bananas, each equally likely to occupy position three. The conditional probability is thereforeThis uses exchangeability of the unconditioned positions; there is no need to assume the first lunch contained a particular fruit.

Each position has apple probability $1/2$. Applying Bayes' theorem to the preceding conditional probability givesThis also agrees with exchangeability of the second and third positions.

By the inclusion-exclusion principle for probabilities,After an initial banana, five of the nine remaining pieces are apples, so the intersection probability is $(1/2)(5/9)=5/18$. ConsequentlyAs a check, the complementary event is an apple followed by a banana, which also has probability $5/18$.

Let $I_j$ be the indicator random variable of an apple in lunch $j$. The number remaining is $R=5-I_1-I_2-I_3$. Each expected value $E[I_j]=P(A_j)=1/2$. The linearity of expectation therefore givesAlthough the indicator random variables are dependent under sampling without replacement, the linearity of expectation needs no independence assumption.

Let $N$ count rotten pieces. Each item supplies an independent Bernoulli trial with success probability $p$, so $N$ has a binomial distribution with parameters $10,p$. There are $\left(\genfrac{}{}{0ex}{}{10}{n}\right)$ choices of the rotten items, and each such choice has probability $p^n(1-p{)}^{10-n}$. HenceFor $p=0$ or $p=1$, interpret this as the corresponding distribution concentrated at zero or ten; no ambiguous $0^0$ evaluation is needed.

Write $N={\sum }_{j=1}^{10}{X}_j$, where the independent Bernoulli random variables indicate rottenness. Since $X_j^2=X_j$, their variances are $E[X_j^2]-E[X_j{]}^2=p-p^2$. The independence makes all cross covariances zero, soThis standard deviation vanishes at $p=0$ and $p=1$, when the count is certain.