Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 16

A uniformly shuffled deck makes all unordered five-card hands equally likely, giving $\left(\genfrac{}{}{0ex}{}{52}{5}\right)$ possible hands. Choose the rank appearing four times in $13$ ways; all four suits of that rank are then fixed. The remaining card can be any of the other $48$ cards. No hand is counted twice, because five cards cannot contain two different complete four-card ranks. This use of combinatorics gives the probabilityHere $\left(\genfrac{}{}{0ex}{}{52}{5}\right)=2598960$ is a binomial coefficient; the probability also simplifies to $1/4165\approx 0.000240096$.

Assume that wrong-note events are independent events and that each note has the same error probability $p$, with $0\le p\le 1$. Then their count $N$ has a binomial distribution. Choose the $n$ mistaken positions in $\left(\genfrac{}{}{0ex}{}{100}{n}\right)$ ways and multiply the probabilities of the errors and correct notes:The endpoint cases $p=0$ or $p=1$ are understood by continuity, giving deterministic counts. More than half means at least $51$ mistakes, soThe single-note error probability alone would not determine this count distribution if mistakes were correlated; independence is the additional standard modeling assumption.

A probability density function must be nonnegative. The printed condition $s\ne 0$ is therefore too weak: for $s<0$ the given prefactor makes the function negative everywhere. The location-scale form of the logistic distribution requires $s>0$, which is assumed in the remaining parts.

Set $t=(x-\mu )/(2s)$, so $dx=2sdt$. The normalization of a probability distribution follows from the antiderivative of the squared hyperbolic secant:For completeness, the same expression with $s<0$ integrates to $-1$, since the transformed limits reverse. If a signed scale were desired, replacing its prefactor by $1/(4|s|)$ would give a valid density, but that is a correction to the printed formula.

For $s>0$, integrate the probability density function to obtain the cumulative distribution functionIf $\mu >0$, the requested probability is $F_X(\mu )-F_X(0)$. Since the hyperbolic tangent is odd, this givesThe second case matters because the question places no sign restriction on $\mu$: the event is empty when its upper endpoint is nonpositive. Using an oriented integral from $0$ to a negative $\mu$ would give a negative number, not a probability.

The same integration by substitution gives the expected valueThe remaining integrand is an odd function. Its integral is absolutely convergent, since ${\mathrm{sech}}^{2}t\le 4e^{-2|t|}$ and $|t|e^{-2|t|}$ is integrable. Thus cancellation under symmetry in integration is legitimate, rather than merely a principal-value calculation. ConsequentlyEquivalently, the logistic distribution is symmetric about $\mu$, and its exponentially decaying tails ensure that its expected value exists.