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2010 Paper 2 Q13
D: 1600.0 B: 1502.2
probability tree diagram optimisation inequality combinatorics strategy comparison complementary counting

Rosalind wants to join the Stepney Chess Club. In order to be accepted, she must play a challenge match consisting of several games against Pardeep (the Club champion) and Quentin (the Club secretary), in which she must win at least one game against each of Pardeep and Quentin. From past experience, she knows that the probability of her winning a single game against Pardeep is \(p\) and the probability of her winning a single game against Quentin is \(q\), where \(0 < p < q < 1\).

  1. The challenge match consists of three games. Before the match begins, Rosalind must choose either to play Pardeep twice and Quentin once or to play Quentin twice and Pardeep once. Show that she should choose to play Pardeep twice.
  2. In order to ease the entry requirements, it is decided instead that the challenge match will consist of four games. Now, before the match begins, Rosalind must choose whether to play Pardeep three times and Quentin once (strategy 1), or to play Pardeep twice and Quentin twice (strategy 2) or to play Pardeep once and Quentin three times (strategy 3). Show that, if \(q-p > \frac 12\), Rosalind should choose strategy 1. If \(q-p<\frac12\), give examples of values of \(p\) and \(q\) to show that strategy 2 can be better or worse than strategy 1.

Solution:

  1. If she plays \(P\) twice her probability is \(q(p^2+2p(1-p)) = qp(2-p)\). If she plays \(Q\) twice her probability is \(pq(2-q)\). Since \(p < q\) she should play \(P\) twice.
  2. Under strategy 1, her probability is \(q(p^3+3p^2(1-p)+3p(1-p)^2) = qp(p^2+3p-3p^2+3-6p+3p^2) = qp(3-3p+p^2)\) Under strategy 2 her probability is \((p^2+2p(1-p))(q^2+2q(1-q)) = pq(2-p)(2-q)\). Under strategy 3 her probability is \(qp(3-3q+q^2)\) \begin{align*} && q - p &> \frac12 \\ \Rightarrow && (2-p)(2-q) & < (2-p)(\frac32 - p) \\ &&&= 3 - \frac72p + p^2 \\ &&&< 3- 3p + p^2 \end{align*} Therefore Strategy 1 dominates if \(q-p > \frac12\). If \(p = \frac14, q = \frac12\) then \((2-p)(2-q) =\frac74 \cdot \frac32 = \frac{21}{8}\) and \(3-3p + p^2 = 3 - \frac34 + \frac1{16} = \frac{48-12+1}{16} = \frac{37}{16} < \frac{42}{16}\) so strategy 2 dominates. Notice that strategy 1 always dominates strategy 3 since \(f(x) = 3-3x+x^2\) is decreasing for \(x < 1.5\). If \(p = \frac14, q = \frac12\) then \((2-p)(2-q) =\frac74 \cdot \frac32 = \frac{21}{8}\) and \(3-3p + p^2 = 3 - \frac34 + \frac1{16} = \frac{48-12+1}{16} = \frac{37}{16} < \frac{42}{16}\) so strategy 2 dominates. For strategy 1 to dominate, we need \(3-3p+p^2 > (2-q)(2-p)\) or \(\frac{3-3p+p^2}{2-p} > 2-q\). When \(p = \frac12\) this is \(\frac{3-\frac32 + \frac14}{2 - \frac12} = \frac{\frac{7}{4}}{\frac{3}{2}} = \frac76 = 2-\frac{5}{6}\) so take any value of \(q\) larger than \(\frac56\).

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2004 Paper 2 Q14
D: 1600.0 B: 1484.0
probability principle of inclusion-exclusion combinatorics complementary counting conditional probability multinomial distribution inequality

Explain why, if \(\mathrm{A, B}\) and \(\mathrm{C}\) are three events, \[ \mathrm{P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) +P(A \cap B \cap C)}, \] where \(\mathrm{P(X)}\) denotes the probability of event \(\mathrm{X}\). A cook makes three plum puddings for Christmas. He stirs \(r\) silver sixpences thoroughly into the pudding mixture before dividing it into three equal portions. Find an expression for the probability that each pudding contains at least one sixpence. Show that the cook must stir 6 or more sixpences into the mixture if there is to be less than \({1 \over 3}\) chance that at least one of the puddings contains no sixpence. Given that the cook stirs 6 sixpences into the mixture and that each pudding contains at least one sixpence, find the probability that there are two sixpences in each pudding.

Solution:

TikZ diagram
When we add everything in \(A\),\(B\), \(C\) we overcount the overlaps. When we remove the overlaps we remove the centre section too many times, so we have to add it back on in the end. Let \(X_i\) be the probability that the \(i\)th pudding contains a sixpence. \begin{align*} && \mathbb{P}(X_1^c \cup X_2^c \cup X_3^c) &=\mathbb{P}(X_1^c \cap X_2^c \cap X_3^c) + \mathbb{P}(X_1^c)+\mathbb{P}(X_2^c)-\mathbb{P}(X_3^c)+\\ &&&\quad\quad-\mathbb{P}(X_1^c \cap X_2^c )-\mathbb{P}( X_2^c \cap X_3^c)-\mathbb{P}(X_1^c \cap X_3^c) \\ &&&= 0 + (\tfrac23)^r+ (\tfrac23)^r+ (\tfrac23)^r + \\ &&&\quad\quad - (\tfrac13)^r- (\tfrac13)^r- (\tfrac13)^r \\ &&&= \frac{3\cdot2^r-3}{3^{r}} \\ \Rightarrow && \mathbb{P}(\text{all contain a sixpence}) &= 1 - \frac{3\cdot2^r-3}{3^{r}} \\ &&&= \frac{3^r-3\cdot2^r+3}{3^r} \end{align*} When \(r = 5\) we have \(\frac{3 \cdot 32-3}{3^5} = \frac{31}{81} > \frac13\) When \(r = 6\) we have \(\frac{3 \cdot 64-3}{3^6} = \frac{7}{27} < \frac13\) Therefore, the chef must stir in at least \(6\). \begin{align*} && \mathbb{P}(\text{two in each}|\text{at least 1 in each}) &= \frac{ \mathbb{P}(\text{two in each} \cap \text{at least 1 in each}) }{ \mathbb{P}(\text{at least 1 in each}) } \\ &&&= \frac{ \mathbb{P}(\text{two in each}) }{ \mathbb{P}(\text{at least 1 in each}) } \\ &&&= \frac{90/3^6}{20/27} \\ &&&= \frac{1/3}{2} = \frac16 \end{align*}

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1998 Paper 1 Q12
D: 1484.0 B: 1606.9
probability independent events combinatorics divisibility integer solutions stars and bars complementary counting

Suppose that a solution \((X,Y,Z)\) of the equation \[X+Y+Z=20,\] with \(X\), \(Y\) and \(Z\) non-negative integers, is chosen at random (each such solution being equally likely). Are \(X\) and \(Y\) independent? Justify your answer. Show that the probability that \(X\) is divisible by \(5\) is \(5/21\). What is the probability that \(XYZ\) is divisible by 5?

Solution: They are not independent: \begin{align*} && \mathbb{P}(X = 20 \,\, \cap Y = 20) = 0 \\ && \mathbb{P}(X = 20 )\mathbb{P}(Y = 20) \neq 0 \\ \end{align*} \begin{align*} X = 0: && 21 \text{ solutions} \\ X = 5: && 16 \text{ solutions} \\ X = 10: && 11 \text{ solutions} \\ X = 15: && 6 \text{ solutions} \\ X = 20: && 1 \text{ solutions} \\ 5 \mid X: && 55 \text{ solutions} \\ \\ && \binom{20+2}{2} = 11 \cdot 21 \text{ total solutions} \\ \Rightarrow && \mathbb{P}(5 \mid X) = \frac{55}{11 \cdot 21} = \frac{5}{21} \end{align*} \begin{align*} \mathbb{P}(5 \mid XYZ) &= 3\cdot \mathbb{P}(5 \mid X) - 2\mathbb{P}(5 \mid X, Y, Z) \\ &= \frac{3 \cdot 55 - 2 \cdot \binom{4+2}{2}}{11 \cdot 21} = \frac{35}{77} \end{align*}

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1998 Paper 3 Q12
D: 1700.0 B: 1482.8
probability independent events complementary counting graph theory tetrahedron connectivity inclusion-exclusion polynomial

The mountain villages \(A,B,C\) and \(D\) lie at the vertices of a tetrahedron, and each pair of villages is joined by a road. After a snowfall the probability that any road is blocked is \(p\), and is independent of the conditions of any other road. The probability that, after a snowfall, it is possible to travel from any village to any other village by some route is \(P\). Show that $$ P =1- p^2(6p^3-12p^2+3p+4). $$ %In the case \(p={1\over 3}\) show that this probability is \({208 \over 243}\).

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1992 Paper 1 Q16
D: 1500.0 B: 1504.2
probability independent events graph theory network reliability road network complementary counting optimisation

The four towns \(A,B,C\) and \(D\) are linked by roads \(AB,AC,CB,BD\) and \(CD.\) The probability that any one road will be blocked by snow on the 1st of January is \(p\), independent of what happens to any other \([0 < p < 1]\). Show that the probability that any open route from \(A\) to \(D\) is \(ABCD\) is \[ p^{2}(1-p)^{3}. \] In order to increase the probability that it is possible to get from \(A\) to \(D\) by a sequence of unblocked roads the government proposes either to snow-proof the road \(AB\) (so that it can never be blocked) or to snow-proof the road \(CB.\) Because of the high cost it cannot do both. Which road should it choose (or are both choices equally advantageous)? In fact, \(p=\frac{1}{10}\) and the government decides that it is only worth going ahead if the present probability of \(A\) being cut off from \(D\) is greater than \(\frac{1}{100}.\) Will it go ahead?

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