Problems

\(47231\) is a five-digit number whose digits sum to \(4+7+2+3+1 = 17\,\).

1. Show that there are \(15\) five-digit numbers whose digits sum to \(43\). You should explain your reasoning clearly.
2. How many five-digit numbers are there whose digits sum to \(39\)?

A pack of cards consists of \(n+1\) cards, which are printed with the integers from \(0\) to \(n\). A~game consists of drawing cards repeatedly at random from the pack until the card printed with 0 is drawn, at which point the game ends. After each draw, the player receives \(\pounds 1\) if the card drawn shows any of the integers from \(1\) to \(w\) inclusive but receives nothing if the card drawn shows any of the integers from \(w+1\) to \(n\) inclusive.

1. In one version of the game, each card drawn is replaced immediately and randomly in the pack. Explain clearly why the probability that the player wins a total of exactly \(\pounds 3\) is equal to the probability of the following event occurring: out of the first four cards drawn which show numbers in the range \(0\) to \(w\), the numbers on the first three are non-zero and the number on the fourth is zero. Hence show that the probability that the player wins a total of exactly \(\pounds 3\) is equal to \(\displaystyle \frac{w^3}{(w+1)^4}\). Write down the probability that the player wins a total of exactly \(\pounds r\) and hence find the expected total win.
2. In another version of the game, each card drawn is removed from the pack. Show that the expected total win in this version is half of the expected total win in the other version.

Three pirates are sharing out the contents of a treasure chest containing \(n\) gold coins and \(2\) lead coins. The first pirate takes out coins one at a time until he takes out one of the lead coins. The second pirate then takes out coins one at a time until she draws the second lead coin. The third pirate takes out all the gold coins remaining in the chest. Find:

1. the probability that the first pirate will have some gold coins;
2. the probability that the second pirate will have some gold coins;
3. the probability that all three pirates will have some gold coins.

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.

Show Solution

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Solution:

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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*}

If a football match ends in a draw, there may be a "penalty shoot-out". Initially the teams each take 5 shots at goal. If one team scores more times than the other, then that team wins. If the scores are level, the teams take shots alternately until one team scores and the other team does not score, both teams having taken the same number of shots. The team that scores wins. Two teams, Team A and Team B, take part in a penalty shoot-out. Their probabilities of scoring when they take a single shot are \(p_A\) and \(p_B\) respectively. Explain why the probability \(\alpha\) of neither side having won at the end of the initial \(10\)-shot period is given by $$\alpha =\sum_{i=0}^5\binom{5}{i}^2(1-p_A)^i(1-p_B)^i\,p_A^{5-i}p_B^{5-i}.$$ Show that the expected number of shots taken is \(\displaystyle 10+ \frac{2\alpha}\beta\;,\) where \(\beta=p_A+p_B-2p_Ap_B\,.\)

Four students, Arthur, Bertha, Chandra and Delilah, exchange gossip. When Arthur hears a rumour, he tells it to one of the other three without saying who told it to him. He decides whom to tell by choosing at random amongst the other three, omitting the ones that he knows have already heard the rumour. When Bertha, Chandra or Delilah hear a rumour, they behave in exactly the same way (even if they have already heard it themselves). The rumour stops being passed round when it is heard by a student who knows that the other three have already heard it. Arthur starts a rumour and tells it to Chandra. By means of a tree diagram, or otherwise, show that the probability that Arthur rehears it is \(3/4\). Find also the probability that Bertha hears it twice and the probability that Chandra hears it twice.

Show Solution

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Solution: Without loss of generality, \(C\) will tell \(B\) about the rumour. If \(B\) tells \(D\) then \(D\) can either tell \(A\) or \(C\) at which point either \(A\) is told or the rumour stops spreading.

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Therefore \(\mathbb{P}(\text{Arthur rehears}) = 3/4\) For the chances Chandra hears it twice, still WLOG, assume she tells B:

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So her chance of hearing it twice is \(\frac12\) The person who hears it 3rd has a \(\frac12\) chance of hearing it twice, but the person who hears if 4th has no chance. Therefore Bertha has a \(\frac14\) chance of hearing it twice.

On the basis of an interview, the \(N\) candidates for admission to a college are ranked in order according to their mathematical potential. The candidates are interviewed in random order (that is, each possible order is equally likely).

1. Find the probability that the best amongst the first \(n\) candidates interviewed is the best overall.
2. Find the probability that the best amongst the first \(n\) candidates interviewed is the best or second best overall.

A bag contains \(b\) black balls and \(w\) white balls. Balls are drawn at random from the bag and when a white ball is drawn it is put aside.

1. If the black balls drawn are also put aside, find an expression for the expected number of black balls that have been drawn when the last white ball is removed.
2. If instead the black balls drawn are put back into the bag, prove that the expected number of times a black ball has been drawn when the first white ball is removed is \(b/w\,\). Hence write down, in the form of a sum, an expression for the expected number of times a black ball has been drawn when the last white ball is removed.

I have \(k\) different keys on my key ring. When I come home at night I try one key after another until I find the key that fits my front door. What is the probability that I find the correct key in exactly \(n\) attempts in each of the following three cases?

1. At each attempt, I choose a key that I have not tried before but otherwise each choice is equally likely.
2. At each attempt, I choose a key from all my keys and each of the \(k\) choices is equally likely.
3. At the first attempt, I choose from all my keys and each of the \(k\) choices is equally likely. Thereafter, I choose from the keys that I did not try the previous time but otherwise each choice is equally likely.

Every person carries two genes which can each be either of type \(A\) or of type \(B\). It is known that \(81\%\) of the population are \(AA\) (i.e. both genes are of type \(A\)), \(18\%\) are \(AB\) (i.e. there is one gene of type \(A\) and one of type \(B\)) and \(1\%\) are \(BB\). A child inherits one gene from each of its parents. If one parent is \(AA\), the child inherits a gene of type \(A\) from that parent; if the parent is \(BB\), the child inherits a gene of type \(B\) from that parent; if the parent is \(AB\), the inherited gene is equally likely to be \(A\) or \(B\).

1. Given that two \(AB\) parents have four children, show that the probability that two of them are \(AA\) and two of them are \(BB\) is \(3/128\).
2. My mother is \(AB\) and I am \(AA\). Find the probability that my father is \(AB\).

A group of biologists attempts to estimate the magnitude, \(N\), of an island population of voles ({\it Microtus agrestis}). Accordingly, the biologists capture a random sample of 200 voles, mark them and release them. A second random sample of 200 voles is then taken of which 11 are found to be marked. Show that the probability, \(p_N\), of this occurrence is given by $$ p_N = k{{{\big((N-200)!\big)}^2} \over {N!(N-389)!}}, $$ where \(k\) is independent of \(N\). The biologists then estimate \(N\) by calculating the value of \(N\) for which \(p_N\) is a maximum. Find this estimate. All unmarked voles in the second sample are marked and then the entire sample is released. Subsequently a third random sample of 200 voles is taken. Write down the probability that this sample contains exactly \(j\) marked voles, leaving your answer in terms of binomial coefficients. Deduce that $$ \sum_{j=0}^{200}{389 \choose j}{3247 \choose {200-j}} = {3636 \choose 200}. $$

In the game of endless cricket the scores \(X\) and \(Y\) of the two sides are such that \[ \P (X=j,\ Y=k)=\e^{-1}\frac{(j+k)\lambda^{j+k}}{j!k!},\] for some positive constant \(\lambda\), where \(j,k = 0\), \(1\), \(2\), \(\ldots\).

1. Find \(\P(X+Y=n)\) for each \(n>0\).
2. Show that \(2\lambda \e^{2\lambda-1}=1\).
3. Show that \(2x \e^{2x-1}\) is an increasing function of \(x\) for \(x>0\) and deduce that the equation in (ii) has at most one solution and hence determine \(\lambda\).
4. Calculate the expectation \(\E(2^{X+Y})\).

How many integers between \(10\,000\) and \(100\,000\) (inclusive) contain exactly two different digits? (\(23\,332\) contains exactly two different digits but neither of \(33\,333\) and \(12\,331\) does.)

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?

Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins. In how many ways can you make up 20 pence using 20p, 10p, 5p, 2p and 1p coins? You are reminded that no credit will be given for unexplained answers.