Year: 2013
Paper: 1
Question Number: 13
Course: LFM Stats And Pure
Section: Discrete Probability Distributions
Around 1500 candidates sat this paper, a significant increase on last year. Overall, responses were good with candidates finding much to occupy them profitably during the three hours of the examination. In hindsight, two or three of the questions lacked sufficient 'punch' in their later parts, but at least most candidates showed sufficient skill to identify them and work on them as part of their chosen selection of questions. On the whole, nearly all candidates managed to attempt 4-6 questions – although there is always a significant minority who attempt 7, 8, 9, … bit and pieces of questions – and most scored well on at least two. Indeed, there were many scripts with 6 question-attempts, most or all of which were fantastically accomplished mathematically, and such excellence is very heart-warming.
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1532.0
Banger Comparisons: 2
From the integers $1, 2, \ldots , 52$, I choose seven (distinct) integers at random, all choices being equally likely. From these seven, I discard any pair that sum to 53. Let $X$ be the random variable the value of which is the number of discarded pairs. Find the probability distribution of $X$ and show that $\E (X) = \frac 7 {17}$.
\textbf{Note:} $7\times 17 \times 47 =5593$.
There are $\binom{26}3\binom{23}{1}2$ ways to obtain $3$ pairs
There are $\binom{26}2 \binom{24}3 \cdot 2^3$ ways to obtain $2$ pairs
There are $\binom{26}1 \binom{25}5 \cdot 2^5$ ways to obtain $1$ pairs
There are $\binom{26}7 \cdot 2^7$ ways to obtain $0$ pairs
There are $\binom{52}{7}$ ways to choose our integers, so
\begin{align*}
&& \mathbb{P}(X = 3) &= \frac{\binom{26}{3} \cdot \binom{23}{1} \cdot 2}{\binom{52}{7}} \\
&&&= \frac{7! \cdot 26 \cdot 25 \cdot 24 \cdot 23 \cdot 2}{3! \cdot 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46} \\
&&&= \frac{7 \cdot 6 \cdot 5 \cdot 4 }{51 \cdot 2\cdot 49 \cdot 2\cdot 47 \cdot 2} \\
&&&= \frac{ 5 }{17\cdot 7\cdot 47} = \frac{5}{5593} \\
\\
&& \mathbb{P}(X = 2) &= \frac{\binom{26}2 \binom{24}3 \cdot 2^3}{\binom{52}{7}} \\
&&&= \frac{220}{5593} \\
\\
&& \mathbb{P}(X = 1) &= \frac{\binom{26}1 \binom{25}5 \cdot 2^5}{\binom{52}{7}} \\
&&&= \frac{1848}{5593} \\
\\
&& \mathbb{P}(X = 0) &= \frac{\binom{26}7 \cdot 2^7}{\binom{52}{7}} \\
&&&= \frac{3520}{5593} \\
\\
&& \mathbb{E}(X) &= \frac{1848}{5593} + 2 \cdot \frac{220}{5593} + 3 \cdot \frac{5}{5593} \\
&&&= \frac{2303}{5593} = \frac{7}{17}
\end{align*}
Notice we can find the expected value directly:
Let $X_i$ be the random variable the $i$th number is discarded.
Notice that $\mathbb{E}(X) = \mathbb{E}\left (\frac12 \left (X_1 +X_2 +X_3 +X_4 +X_5 +X_6 +X_7\right) \right)$ and also notice that each $X_i$ has the same distribution (although not independent!).
Then
\begin{align*}
&&\mathbb{E}(X) &= \frac72 \mathbb{E}(X_i) \\
&&&= \frac72 \cdot \left (1 - \frac{50}{51} \cdot \frac{49}{50} \cdots \frac{45}{46} \right) = \frac74 \left ( 1 - \frac{45}{51}\right) \\
&&&= \frac72 \cdot \frac{6}{51} \\
&&&= \frac7{17}
\end{align*}
After Q10, this was the least popular question on the paper, and supplied the poorest average score on the whole paper of only 2½/20. I have little doubt that the principal reason for both these factors is the lack of any helpful structure or given answers within the question. Essentially, this problem is that of the set-up for a game of Solitaire, but stripped of its context. In this game, when a standard pack of playing cards, suitably shuffled, is laid out at the start, there are seven piles of cards, and each pile has its final card face up. This particular question is looking for cards of the same colour (red or black) and denomination (number or J, Q, K and Ace), giving the 26 pairs. This was, of course, entirely by-the-by as far as candidates were concerned. Unfortunately, most attempts at this question were abandoned very early on as candidates realised they didn't really know what to do. Surprisingly, very few even took the trouble to note that the defined discrete random variable X could only take the values 0, 1, 2 or 3. Following attempts to work out the probability for any these outcomes almost invariably consisted of a jumble of fractions and factorials but without any obvious plan to them, and certainly without any explanatory indicators as to what might actually be intended. Only P(X = 0), being the easiest of the four cases to evaluate, was calculated with any degree of success by any of the candidates who attempted the question.