Problems

A bag contains three coins. The probabilities of their showing heads when tossed are \(p_1\), \(p_2\) and \(p_3\).

1. A coin is taken at random from the bag and tossed. What is the probability that it shows a head?
2. A coin is taken at random from the bag (containing three coins) and tossed; the coin is returned to the bag and again a coin is taken at random from the bag and tossed. Let \(N_1\) be the random variable whose value is the number of heads shown on the two tosses. Find the expectation of \(N_1\) in terms of \(p\), where \(p = \frac{1}{3}(p_1+p_2+p_3)\,\), and show that \(\var(N_1) =2p(1-p)\,\).
3. Two of the coins are taken at random from the bag (containing three coins) and tossed. Let \(N_2\) be the random variable whose value is the number of heads showing on the two coins. Find \(\E(N_2)\) and \(\var(N_2)\).
4. Show that \(\var(N_2)\le \var(N_1)\), with equality if and only if \(p_1=p_2=p_3\,\).

Four children, \(A\), \(B\), \(C\) and \(D\), are playing a version of the game `pass the parcel'. They stand in a circle, so that \(ABCDA\) is the clockwise order. Each time a whistle is blown, the child holding the parcel is supposed to pass the parcel immediately exactly one place clockwise. In fact each child, independently of any other past event, passes the parcel clockwise with probability \(\frac{1}{4}\), passes it anticlockwise with probability \(\frac{1}{4}\) and fails to pass it at all with probability \(\frac{1}{2}\). At the start of the game, child \(A\) is holding the parcel. The probability that child \(A\) is holding the parcel just after the whistle has been blown for the \(n\)th time is \(A_n\), and \(B_n\), \(C_n\) and \(D_n\) are defined similarly.

1. Find \(A_1\), \(B_1\), \(C_1\) and \(D_1\). Find also \(A_2\), \(B_2\), \(C_2\) and \(D_2\).
2. By first considering \(B_{n+1}+D_{n+1}\), or otherwise, find \(B_n\) and \(D_n\). Find also expressions for \(A_n\) and \(C_n\) in terms of \(n\).

I have a sliced loaf which initially contains \(n\) slices of bread. Each time I finish setting a STEP question, I make myself a snack: either toast, using one slice of bread; or a sandwich, using two slices of bread. I make toast with probability \(p\) and I make a sandwich with probability \(q\), where \(p+q=1\), unless there is only one slice left in which case I must, of course, make toast. Let \(s_r\) (\(1 \le r \le n\)) be the probability that the \(r\)th slice of bread is the second of two slices used to make a sandwich and let \(t_r\) (\(1 \le r \le n\)) be the probability that the \(r\)th slice of bread is used to make toast. What is the value of \(s_1\)? Explain why the following equations hold: \begin{align*} \phantom{\hspace{2cm} (2\le r \le n-1)} t_r &= (s_{r-1}+ t_{r-1})\,p \hspace{2cm} (2\le r \le n-1)\,; \\ \phantom{\hspace{1.53cm} (2\le r \le n) } s_r &= 1- (s_{r-1} + t_{r-1}) \hspace{1.53cm} ( 2\le r \le n )\,. \end{align*} Hence, or otherwise, show that \(s_{r} = q(1-s_{r-1})\) for \(2\le r\le n-1\). Show further that \[ \phantom{\hspace{2.7cm} (1\le r\le n)\,,} s_r = \frac{q+(-q)^r}{1+q} \hspace{2.7cm} (1\le r\le n-1)\,, \, \hspace{0.14cm} \] and find the corresponding expression for \(t_r\). Find also expressions for \(s_n\) and \(t_n\) in terms of \(q\).

Four players \(A\), \(B\), \(C\) and \(D\) play a coin-tossing game with a fair coin. Each player chooses a sequence of heads and tails, as follows: Player A: HHT; Player B: THH; Player C: TTH; Player D: HTT. The coin is then tossed until one of these sequences occurs, in which case the corresponding player is the winner.

1. Show that, if only \(A\) and \(B\) play, then \(A\) has a probability of \(\frac14\) of winning.
2. If all four players play together, find the probabilities of each one winning.
3. Only \(B\) and \(C\) play. What is the probability of \(C\) winning if the first two tosses are TT? Let the probabilities of \(C\) winning if the first two tosses are HT, TH and HH be \(p\), \(q\) and \(r\), respectively. Show that \(p=\frac12 +\frac12q\). Find the probability that \(C\) wins.

A modern villa has complicated lighting controls. In order for the light in the swimming pool to be on, a particular switch in the hallway must be on and a particular switch in the kitchen must be on. There are four identical switches in the hallway and four identical switches in the kitchen. Guests cannot tell whether the switches are on or off, or what they control. Each Monday morning a guest arrives, and the switches in the hallway are either all on or all off. The probability that they are all on is \(p\) and the probability that they are all off is \(1-p\). The switches in the kitchen are each on or off, independently, with probability \(\frac12\).

1. On the first Monday, a guest presses one switch in the hallway at random and one switch in the kitchen at random. Find the probability that the swimming pool light is on at the end of this process. Show that the probability that the guest has pressed the swimming pool light switch in the hallway, given that the light is on at the end of the process, is \(\displaystyle \frac{1-p}{1+2p}\).
2. On each of seven Mondays, guests go through the above process independently of each other, and each time the swimming pool light is found to be on at the end of the process. Given that the most likely number of days on which the swimming pool light switch in the hallway was pressed is 3, show that \(\frac14 < p < \frac{5}{14}\).

Xavier and Younis are playing a match. The match consists of a series of games and each game consists of three points. Xavier has probability \(p\) and Younis has probability \(1-p\) of winning the first point of any game. In the second and third points of each game, the player who won the previous point has probability \(p\) and the player who lost the previous point has probability \(1-p\) of winning the point. If a player wins two consecutive points in a single game, the match ends and that player has won; otherwise the match continues with another game.

1. Let \(w\) be the probability that Younis wins the match. Show that, for \(p\ne0\), \[ w = \frac{1-p^2}{2-p}. \] Show that \(w>\frac12\) if \(p<\frac12\), and \(w<\frac12\) if \(p>\frac12\). Does \(w\) increase whenever \(p\) decreases?
2. If Xavier wins the match, Younis gives him \(\pounds1\); if Younis wins the match, Xavier gives him \(\pounds k\). Find the value of \(k\) for which the game is `fair' in the case when \(p =\frac23\).
3. What happens when \(p = 0\)?

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.

Prove that, for any real numbers \(x\) and \(y\), \(x^2+y^2\ge2xy\,\).

1. Carol has two bags of sweets. The first bag contains \(a\) red sweets and \(b\) blue sweets, whereas the second bag contains \(b\) red sweets and \(a\) blue sweets, where \(a\) and \(b\) are positive integers. Carol shakes the bags and picks one sweet from each bag without looking. Prove that the probability that the sweets are of the same colour cannot exceed the probability that they are of different colours.
2. Simon has three bags of sweets. The first bag contains \(a\) red sweets, \(b\) white sweets and \(c\) yellow sweets, where \(a\), \(b\) and \(c\) are positive integers. The second bag contains \(b\) red sweets, \(c\) white sweets and \(a\) yellow sweets. The third bag contains \(c\) red sweets, \(a\) white sweets and \(b\) yellow sweets. Simon shakes the bags and picks one sweet from each bag without looking. Show that the probability that exactly two of the sweets are of the same colour is \[ \frac {3(a^2b+b^2c+c^2a+ab^2 + bc^2 +ca^2)}{(a+b+c)^3}\,, \] and find the probability that the sweets are all of the same colour. Deduce that the probability that exactly two of the sweets are of the same colour is at least 6 times the probability that the sweets are all of the same colour.

Satellites are launched using two different types of rocket: the Andover and the Basingstoke. The Andover has four engines and the Basingstoke has six. Each engine has a probability~\(p\) of failing during any given launch. After the launch, the rockets are retrieved and repaired by replacing some or all of the engines. The cost of replacing each engine is \(K\). For the Andover, if more than one engine fails, all four engines are replaced. Otherwise, only the failed engine (if there is one) is replaced. Show that the expected repair cost for a single launch using the Andover is \[ 4Kp(1+q+q^2-2q^3) \ \ \ \ \ \ \ \ \ \ \ \ \ (q=1-p) \tag{*} \] For the Basingstoke, if more than two engines fail, all six engines are replaced. Otherwise only the failed engines (if there are any) are replaced. Find, in a form similar to \((*)\), the expected repair cost for a single launch using the Basingstoke. Find the values of \(p\) for which the expected repair cost for the Andover is \(\frac23\) of the expected repair cost for the Basingstoke.

Bag \(P\) and bag \(Q\) each contain \(n\) counters, where \(n\ge2\). The counters are identical in shape and size, but coloured either black or white. First, \(k\) counters (\(0\le k\le n\)) are drawn at random from bag \(P\) and placed in bag \(Q\). Then, \(k\) counters are drawn at random from bag \(Q\) and placed in bag \(P\).

1. If initially \(n-1\) counters in bag \(P\) are white and one is black, and all \(n\) counters in bag \(Q\) are white, find the probability in terms of \(n\) and \(k\) that the black counter ends up in bag \(P\). Find the value or values of \(k\) for which this probability is maximised.
2. If initially \(n-1\) counters in bag \(P\) are white and one is black, and \(n-1\) counters in bag \(Q\) are white and one is black, find the probability in terms of \(n\) and \(k\) that the black counters end up in the same bag. Find the value or values of \(k\) for which this probability is maximised.

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.

A pack of \(2n\) (where \(n\geqslant4\)) cards consists of two each of \(n\) different sorts. If four cards are drawn from the pack without replacement show that the probability that no pairs of identical cards have been drawn is \[ \frac{4(n-2)(n-3)}{(2n-1)(2n-3)}. \] Find the probability that exactly one pair of identical cards is included in the four. If \(k\) cards are drawn without replacement and \(2 < k < 2n,\) find an expression for the probability that there are exactly \(r\) pairs of identical cards included when \(r < \frac{1}{2}k.\) For even values of \(k\) show that the probability that the drawn cards consist of \(\frac{1}{2}k\) pairs is \[ \frac{1\times3\times5\times\cdots\times(k-1)}{(2n-1)(2n-3)\cdots(2n-k+1)}. \]