Year: 2024
Paper: 3
Question Number: 11
Course: LFM Stats And Pure
Section: Binomial Distribution
No solution available for this problem.
The total entry was an increase on that of 2023 by more than 10%. One question was attempted by more than 98% of candidates, another two by about 80%, and another five by between 50% and 70%. The remaining four questions were attempted by between 5% and 30% of candidates, these being from Section B: Mechanics, and Section C: Probability and Statistics, though the Statistics questions were in general attempted more often and more successfully. All questions were perfectly solved by some candidates. About 84% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, you may use without proof the results
\[\sum_{r=0}^{n} \binom{n}{r} = 2^n \quad \text{and} \quad \sum_{r=0}^{n} r\binom{n}{r} = n\,2^{n-1}.\]
\begin{questionparts}
\item Show that
\[r\binom{2n}{r} = (2n+1-r)\binom{2n}{2n+1-r}\]
for $1 \leqslant r \leqslant 2n$. Hence show that
\[\sum_{r=0}^{2n} r\binom{2n}{r} = 2\sum_{r=n+1}^{2n} r\binom{2n}{r}.\]
\item A fair coin is tossed $2n$ times. The value of the random variable $X$ is whichever is the larger of the number of heads and the number of tails shown. If $n$ heads and $n$ tails are shown, then $X = n$.
Show that
\[\mathrm{E}(X) = n\left(1 + \frac{1}{2^{2n}}\binom{2n}{n}\right).\]
\item Show that $\dfrac{1}{2^{2n}}\dbinom{2n}{n}$ decreases as $n$ increases.
\item In a game, you choose a value of $n$ and pay $\pounds n$; then a fair coin is tossed $2n$ times. You win an amount in pounds equal to the larger of the number of heads and the number of tails shown. If $n$ heads and $n$ tails are shown, then you win $\pounds n$. How should you choose $n$ to maximise your expected winnings per pound paid?
\end{questionparts}
Very nearly 30% of the candidates attempted this, making it the most popular non-Pure question, and they did so relatively successfully with a mean score of nearly 11/20, better than all but question 1. A significant number of candidates gained full or close to full credit. Part (i) was generally well executed, although using r(2n choose r) = (2n+1−r)(2n choose 2n+1−r) for r = 0 without justification was a common error. In part (ii), a common error was using an incorrect probability distribution for the random variable X, common examples included asserting that X itself was binomially distributed as B(2n, 1/2), or asserting that either P(X = k) = (1/2^2n)(2n choose k) or P(X = k) = (2/2^2n)(2n choose k) for all n ≤ k ≤ 2n. Showing that (1/2^2n)(2n choose n) is a decreasing function of n in part (iii) was generally well executed; a few students considered the difference between (1/2^2n)(2n choose n) and (1/2^(2n+2))(2n+2 choose n+1), rather than the ratio, which lead to a largely similar, but slightly more involved, computation. Part (iv) commonly saw candidates trying to maximise total expected winnings, rather than expected winnings per pound. However generally the standard of responses to this question was quite high.