A train has \(n\) seats, where \(n \geqslant 2\). For a particular journey, all \(n\) seats have been sold, and each of the \(n\) passengers has been allocated a seat.
The passengers arrive one at a time and are labelled \(T_1, \ldots, T_n\) according to the order in which they arrive: \(T_1\) arrives first and \(T_n\) arrives last. The seat allocated to \(T_r\) (\(r = 1, \ldots, n\)) is labelled \(S_r\).
Passenger \(T_1\) ignores their allocation and decides to choose a seat at random (each of the \(n\) seats being equally likely). However, for each \(r \geqslant 2\), passenger \(T_r\) sits in \(S_r\) if it is available or, if \(S_r\) is not available, chooses from the available seats at random.
- Let \(P_n\) be the probability that, in a train with \(n\) seats, \(T_n\) sits in \(S_n\). Write down the value of \(P_2\) and find the value of \(P_3\).
- Explain why, for \(k = 2, 3, \ldots, n-1\),
\[
\mathrm{P}\bigl(T_n \text{ sits in } S_n \mid T_1 \text{ sits in } S_k\bigr) = P_{n-k+1},
\]
and deduce that, for \(n \geqslant 3\),
\[
P_n = \frac{1}{n}\Biggl(1 + \sum_{r=2}^{n-1} P_r\Biggr).
\]
- Give the value of \(P_n\) in its simplest form and prove your result by induction.
- Let \(Q_n\) be the probability that, in a train with \(n\) seats, \(T_{n-1}\) sits in \(S_{n-1}\). Determine \(Q_n\) for \(n \geqslant 2\).
A coin is tossed repeatedly. The probability that a head appears is \(p\) and the probability that a tail appears is \(q = 1 - p\).
- A and B play a game. The game ends if two successive heads appear, in which case A wins, or if two successive tails appear, in which case B wins.
Show that the probability that the game never ends is \(0\).
Given that the first toss is a head, show that the probability that A wins is \(\dfrac{p}{1 - pq}\).
Find and simplify an expression for the probability that A wins.
- A and B play another game. The game ends if three successive heads appear, in which case A wins, or if three successive tails appear, in which case B wins.
Show that
\[\mathrm{P}(\text{A wins} \mid \text{the first toss is a head}) = p^2 + (q + pq)\,\mathrm{P}(\text{A wins} \mid \text{the first toss is a tail})\]
and give a similar result for \(\mathrm{P}(\text{A wins} \mid \text{the first toss is a tail})\).
Show that
\[\mathrm{P}(\text{A wins}) = \frac{p^2(1-q^3)}{1-(1-p^2)(1-q^2)}.\]
- A and B play a third game. The game ends if \(a\) successive heads appear, in which case A wins, or if \(b\) successive tails appear, in which case B wins, where \(a\) and \(b\) are integers greater than \(1\).
Find the probability that A wins this game.
Verify that your result agrees with part (i) when \(a = b = 2\).
A multiple-choice test consists of five questions. For each question, \(n\) answers are given (\(n\ge2\)) only one of which is correct and candidates either attempt the question by choosing one of the \(n\) given answers or do not attempt it.
For each question attempted, candidates receive two marks for the correct answer and lose one mark for an incorrect answer. No marks are gained or lost for questions that are not attempted. The pass mark is five.
Candidates A, B and C don't understand any of the questions
so, for any question which they attempt, they each choose one of the \(n\) given answers at random, independently of their choices for any other question.
- Candidate A chooses in advance to attempt exactly \(k\) of the five questions, where \(k=0, 1, 2, 3, 4\) or \(5\). Show that, in order to have the greatest probability of passing the test, she should choose \(k=4\,\).
- Candidate B chooses at random the number of questions he will attempt, the six possibilities being equally likely. Given that Candidate B passed the test find, in terms of \(n\), the probability that he attempted exactly four questions.
[Not on original test: Show that this probability is an increasing function of \(n\).]
- For each of the five questions Candidate C decides whether to attempt the question by tossing a biased coin. The coin has a probability of \(\frac n{n+1}\) of showing a head, and she attempts the question if it shows a head. Find the probability, in terms of \(n\), that Candidate C passes the test.
Show Solution
- Her probability of passing if she answers \(k \leq 2\) is \(0\), since she can attain at most \(4\) marks.
If she attempts \(3\) questions, she needs to get all of them right, hence \(\mathbb{P}(\text{gets all }3\text{ correct}) = \frac{1}{n^3}\).
If she attempts \(4\) questions, we can afford to get one wrong
\begin{align*}
&& \mathbb{P}(\text{passes}|\text{attempts }4) &=\mathbb{P}(4/4) +\mathbb{P}(3/4) \\
&&&= \frac{1}{n^4} + 4\cdot\frac{1}{n^3} \cdot \frac{n-1}{n} \\
&&&= \frac{4n-3}{n^4}
\end{align*}
If she attempts \(5\) questions she can get \(5\) right (10), \(4\) right, \(1\) wrong (7), but \(3\) right will not work (\(6 - 2 = 4< 5\)), hence:
\begin{align*}
&& \mathbb{P}(\text{passes}|\text{attempts }5) &=\mathbb{P}(5/5) +\mathbb{P}(4/5) \\
&&&= \frac{1}{n^5} + 5\cdot\frac{1}{n^4} \cdot \frac{n-1}{n} \\
&&&= \frac{5n-4}{n^5}
\end{align*}
If \(4n-3 > n \Leftrightarrow n \geq 2\) then \(4\) attempts is better than \(3\). If \(4n^2-3n > 5n-4 \Leftrightarrow 4n^2-8n+4 = 4(n-1)^2 > 0 \Leftrightarrow n \geq\) then \(4\) is better than \(5\), but \(n\) is \(\geq 2\) so, \(4\) is the best option.
- \(\,\)
\begin{align*}
&& \mathbb{P}(\text{passes}) &= \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot \frac1{n^3} + \frac16 \frac{4n-3}{n^4} + \frac16 \frac{5n-4}{n^5} \\
&&&= \frac{n^2+4n^2-3n+5n-4}{6n^5} \\
&&&= \frac{5n^2+2n-4}{6n^5} \\
&& \mathbb{P}(\text{answered }4|\text{passes}) &= \frac{\mathbb{P}(\text{answered }4\text{ and passes})}{ \mathbb{P}(\text{passes})} \\
&&&= \frac{\frac16 \cdot \frac{4n-3}{n^4}}{\frac{5n^2-2n-4}{6n^5} } \\
&&&= \frac{4n^2-3n}{5n^2+2n-4}
\end{align*}
Notice that the function takes all values for \(n\) between the roots of the denominator (which are either side of \(0\) and below \(3/4\). Therefore after \(3/4\) the function must be increase since otherwise we would have a quadratic equation with more than \(2\) roots.
- \(\,\) \begin{align*}
&&\mathbb{P}(C \text{ passes}) &= \binom{5}{3} \left ( \frac{n}{n+1} \right)^3 \left ( \frac{1}{n+1}\right)^2 \frac{1}{n^3} + \binom{5}{4} \left ( \frac{n}{n+1} \right)^4 \left ( \frac{1}{n+1}\right) \frac{4n-3}{n^4} +\\
&&&\quad \quad + \binom{5}{5} \left ( \frac{n}{n+1} \right)^5 \frac{5n-4}{n^5} \\
&&&= \frac{10}{(n+1)^5} + \frac{5(4n-3)}{(n+1)^5} + \frac{(5n-4)}{(n+1)^5} \\
&&&= \frac{10+20n-15+5n-4}{(n+1)^5}\\
&&&= \frac{25n-9}{(n+1)^5}\\
\end{align*}
A bag contains eleven small discs, which are identical except that six of the discs are blank and five of the discs are numbered, using the numbers 1, 2, 3, 4 and 5.
The bag is shaken, and four discs are taken one at a time without replacement.
Calculate the probability that:
- all four discs taken are numbered;
- all four discs taken are numbered,
given that the disc numbered ``3'' is taken first;
- exactly two numbered discs are taken,
given that the disc numbered ``3'' is taken first;
- exactly two numbered discs are taken,
given that the disc numbered ``3'' is taken;
- exactly two numbered discs are taken,
given that a numbered disc is taken first;
- exactly two numbered discs are taken,
given that a numbered disc is taken.
Show Solution
There are many ways to do the counting in each question, possibly the clearest way is to always consider the order in which discs are taken, although all methods should work equally well. For some examples Bayes rule also offers a fast solution.
- There are we are choose \(4\) objects in order from \(5\) (ie \({^5\P_4}\)) to obtain valid draws, this is out of a total of picking \(4\) objects from \(11\) (\({^{11}\P_4}\)). Ie the probability is: \(\displaystyle \frac{{^5\P_4}}{{^{11}\P_4}} = \frac{5! \cdot 7!}{11!} = \frac{1}{66}\)
Alternatively, there are \(\binom{5}{4}\) ways to choose four numbered discs, out of \(\binom{11}{4}\) ways to choose four discs. ie \(\displaystyle \binom{5}{4} \Big / \binom{11}{4} = \frac{5 \cdot 4! \cdot 7!}{11!} = \frac{5 \cdot 4 \cdot 3 \cdot 2}{11 \cdot 10 \cdot 9 \cdot 8} = \frac1{66}\)
- \begin{align*}
\mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\
&= \frac{^4\P_3 \big / {^{11}\P_4}}{1/11} \\
&= 11\cdot \frac{4!}{1!} \Bigg / \frac{11!}{7!} \\
&= \frac{4! \cdot 7! \cdot 11}{11!} \\
&= \frac{4\cdot 3 \cdot 2}{10 \cdot 9 \cdot 8} \\
&= \frac{1}{30}
\end{align*}
Alternatively,
\begin{align*}
\mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\
&= \frac{\binom{4}{3} \Bigg / 11 \cdot \binom{10}{3}}{1/11} \\
&= \frac{1}{30}
\end{align*}
Where we are calculating this as "choose one number", then "choose 3 more", which can happen ending up with 3, number, number, number in \(\binom{4}{3}\) ways, and there are \(11 \cdot \binom{10}{3}\) was overall.
Another alternative using Bayes rule:
\begin{align*}
\mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \mathbb{P}( \text{first disc is 3} | \text{all four discs are numbered}) \frac{ \mathbb{P}( \text{all four discs are numbered} }{ \mathbb{P}( \text{first disc is 3} )} \\
&= \frac{\frac{1}{5} \cdot \frac{1}{66}}{\frac{1}{11}} \\
&= \frac1{30}
\end{align*}
- \begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\
&= \frac{3 \cdot {^4\P_1} \cdot {^{6}\P_2} \big / {^{11}\P_4}}{\frac1{11} } \\
&= \frac12
\end{align*}
Alternatively,
\begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\
&= \frac{\binom{4}{1}\binom{6}{2} \Big / 11 \cdot \binom{10}{3}}{1/11} \\
&= \frac12
\end{align*}
- \begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and 3 taken})}{\mathbb{P}( \text{3 taken})} \\
&= \frac{\binom{4}{1}\binom{6}{2} \Big / \binom{11}{4}}{\frac{4}{11}} \\
&= \frac{\frac{2}{11}}{\frac4{11}} \\
&= \frac12
\end{align*}
Using Bayes rule:
\(\mathbb{P}( \text{3 taken}) = \frac{1}{11} + \frac{10}{11}\frac{1}{10} + \frac{10}{11}\frac{9}{10}\frac{1}{9} + \frac{10}{11}\frac{9}{10}\frac89\frac18 = \frac{4}{11}\)
\begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{3 taken | exactly two discs are numbered})\mathbb{P}(\text{exactly two discs are numbered})}{\mathbb{P}( \text{3 taken})} \\
&= \frac{\frac{4}{10} \cdot \binom{5}{2} \binom{6}{2} \Big / \binom{11}{4}}{4/11} \\
&= \frac{\frac4{10}{5 / 11}}{4/11} \\
&= \frac{1}{2}
\end{align*}
- \begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc first}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc first})}{\mathbb{P}( \text{numbered disc first})} \\
&= \frac{3 \cdot {^5\P_1}\cdot{^4\P_1}\cdot{^6\P_2} \Big / {^{11}\P_4}}{\frac{5}{11}} \\
&= \frac{1}{2}
\end{align*}
- \begin{align*}
\mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc taken})}{\mathbb{P}(\text{numbered disc taken})} \\
&= \frac{\mathbb{P}(\text{exactly two discs are numbered})}{1 - \mathbb{P}(\text{not numbered discs taken})} \\
&= \frac{\binom{5}{2}\binom{6}{2} \Big / \binom{11}{4}}{1 - \binom{6}{4} \Big / \binom{11}{4}} \\
&= \frac{\frac{5}{11}}{\frac{21}{22}} \\
&= \frac{10}{21} \neq \frac12
\end{align*}
Oxtown and Camville are connected by three roads,
which are at risk of being blocked by flooding.
On two of the three roads there are two sections
which may be blocked. On the third road
there is only one section which may be blocked.
The probability that each section is blocked is \(p\).
Each section is blocked independently of the other four sections.
Show that the probability that Oxtown is cut off from
Camville is \(p^3 \l 2-p \r^2\).
I want to travel from Oxtown to Camville. I choose
one of the three roads at random and find that my road is not blocked.
Find the probability that I would not have reached Camville
if I had chosen either of the other two roads.
You should factorise your answer as fully as possible.
Comment briefly on the value of this probability in the limit \(p\to1\).
I know that
ice-creams come in \(n\) different sizes, but I don't know what the sizes are.
I am offered one of each in
succession, in random order.
I am certainly going to choose one - the bigger
the better - but I
am not allowed more than one. My strategy is to reject the first
ice-cream I am offered
and choose the first one thereafter that is bigger than the first
one I was offered; if the first ice-cream offered is in fact the biggest one,
then I have to put up with the last one, however small.
Let \(\P_n(k)\) be the probability that I choose the \(k\)th biggest ice-cream,
where \(k=1\) is the biggest and \(k=n\) is the smallest.
- Show that \(\P_4(1) = \frac{11}{24}\) and find \(\P_4(2)\), \(\P_4(3)\)
and \(\P_4(4)\).
- Find an expression for \(\P_n(1)\).
The twins Anna and Bella share a computer and never sign their e-mails.
When I e-mail them, only the twin
currently online responds. The
probability that it is Anna who is online is \(p\) and she answers each
question I ask her truthfully with probability \(a\), independently of all her
other answers, even if a question is repeated. The probability that it is
Bella who is online is~\(q\), where \(q=1-p\), and she answers each question
truthfully with probability \(b\), independently of all her other answers,
even if a question is repeated.
- I send the twins the e-mail:
`Toss a fair coin and answer the following question.
Did the coin come down heads?'. I receive the answer `yes'.
Show that the probability that the coin
did come down heads is \(\frac{1}{2}\) if and
only if \(2(ap+bq)=1\).
- I send the twins the e-mail:
`Toss a fair coin and answer the following question.
Did the coin come down heads?'. I receive the answer `yes'.
I then send the e-mail: `Did the coin come down heads?' and I receive
the answer `no'. Show that the probability (taking into
account these answers) that the coin did come down heads is \(\frac{1}{2}\,\).
- I send the twins the e-mail: `Toss a fair coin and answer the following
question. Did the coin come down heads?'. I receive the answer `yes'.
I then send the e-mail: `Did the coin come down heads?' and I receive
the answer `yes'. Show that, if \(2(ap+bq)=1\),
the probability (taking into account these answers) that the coin did
come down heads is \(\frac{1}{2}\,\).
In a rabbit warren, underground chambers
\(A, B, C\) and \(D\) are at the vertices of a square,
and burrows join \(A\) to \(B\), \ \(B\) to \(C\), \ \(C\) to \(D\) and \(D\) to \(A\).
Each of the chambers also has a tunnel to the surface.
A rabbit finding itself in any chamber runs along one
of the two burrows to a neighbouring chamber, or
leaves the burrow through the tunnel to the surface.
Each of these three possibilities is equally likely.
Let \(p_A\,\), \(p_B\,\), \(p_C\) and \(p_D\) be the probabilities
of a rabbit leaving the burrow through the tunnel from chamber \(A\),
given that it is currently in chamber \(A, B, C\) or \(D\), respectively.
- Explain why \(p_A = \frac13 + \frac13p_B + \frac13 p_D\).
- Determine \(p_A\,\).
- Find the probability
that a rabbit which starts in chamber \(A\) does not visit chamber~\(C\),
given that it eventually leaves the burrow through the tunnel in chamber \(A\).
The random variable \(U\) takes the values \(+1\), \(0\) and \(-1\,\), each with probability \(\frac13\,\). The random variable \(V\) takes the values \(+1\) and \(-1\) as follows:
| if \(U=1\,\), | then \(\P(V=1)= \frac13\) and \(\P(V=-1)=\frac23\,\); |
| if \(U=0\,\), | then \(\P(V=1)= \frac12\) and \(\P(V=-1)=\frac12\,\); |
| if \(U=-1\,\), | then \(\P(V=1)= \frac23\) and \(\P(V=-1)=\frac13\,\). |
- Show that the probability that both roots of the equation \(x^2+Ux+V=0\) are real is \(\frac12\;\).
- Find the expected value of the larger root of the equation \(x^2+Ux+V=0\,\), given that both roots are real.
- Find the probability that the roots of the equation
$$x^3+(U-2V)x^2+(1-2UV)x + U=0$$ are all positive.
Show Solution
- \(\,\) \begin{align*}
&& \mathbb{P}(\text{both roots real}) &= \mathbb{P}(\Delta \geq 0) \\
&&&= \mathbb{P}(U^2 \geq 4V) \\
&&&= \mathbb{P}(V = -1) \\
&&&= \tfrac13 ( \tfrac23 + \tfrac12 + \tfrac13) \\
&&&= \tfrac13 \cdot \tfrac 32 = \frac12
\end{align*}
- Our equations will be:
\(x^2+x-1 = 0\) with larger root \(\frac{-1 + \sqrt{5}}{2}\)
\(x^2-1 = 0\) with larger root \(1\)
\(x^2-x-1 = 0\) with larger root \(\frac{1 + \sqrt5}{2}\)
and the expected value is \begin{align*}
&& \E[\text{larger root}|\text{both real}] &= \frac23 \left ( \frac23
\cdot \frac{-1+\sqrt5}{2} + \frac12 \cdot 1 + \frac13 \cdot \frac{1+\sqrt5}{2} \right) \\
&&&= \frac23 \left ( \frac{2+3\sqrt5}{6} \right) \\
&&&= \frac{2+3\sqrt5}{9}
\end{align*}
- Suppose we have \(x^3+(U-2V)x^2+(1-2UV)x + U = 0\), then for all roots to be positive, we need \(U < 0 \Rightarrow U = -1\) (otherwise there is a root at or below zero).
Therefore our two possible cubics are:
\(x^3 -3x^2+3x-1 = (x-1)^3\) (all roots positive)
\(x^3+x^2-x-1 = (x-1)(x+1)^2\) (not all roots positive!)
Therefore the probability is \(\frac13 \cdot \frac23 = \frac29\)
In a game, a player tosses a biased coin
repeatedly until two successive tails occur, when the game terminates.
For each head which occurs the player wins \(\pounds 1\).
If \(E\) is the expected number of tosses of the
coin in the course of a game, and \(p\) is the probability of a head, explain why
\[
E = p \l 1 + E \r + \l 1 - p \r p \l 2 + E \r + 2 \l 1 - p \r ^2\,,
\]
and hence determine \(E\) in terms of \(p\).
Find also, in terms of \(p\), the expected winnings in the course of a game.
A second game is played,
with the same rules, except that the player continues to
toss the coin until \(r\) successive tails occur.
Show that the expected number of tosses in the
course of a game is given by the expression
\(\displaystyle {1 - q^r \over p q^r}\,\), where \(q = 1 - p\).
In a game for two players, a fair coin is tossed repeatedly. Each player
is assigned a sequence of heads and tails and the player whose sequence appears first wins.
Four players, \(A\), \(B\), \(C\) and \(D\) take turns to play the game.
Each time they play, \(A\) is assigned the sequence
TTH (i.e.~Tail then Tail then Head), \(B\) is assigned THH,
\(C\) is assigned HHT and \(D\) is assigned~HTT.
- \(A\) and \(B\) play the game.
Let
\(p_{\mathstrut\mbox{\tiny HH}}\),
\(p_{\mathstrut\mbox{\tiny HT}}\),
\(p_{\mathstrut\mbox{\tiny TH}}\) and
\(p_{\mathstrut\mbox{\tiny TT}}\)
be the probabilities of \(A\) winning the
game given that the first two tosses of the coin show HH, HT, TH and TT, respectively.
Explain why
\(p_{\mathstrut\mbox{\tiny TT}} = 1\,\),
and why
$p_{\mathstrut\mbox{\tiny HT}} = {1 \over 2} \,
p_{\mathstrut\mbox{\tiny TH}} +
{1\over 2} \,
p_{\mathstrut\mbox{\tiny TT}}\,$.
Show that
$p_{\mathstrut\mbox{\tiny HH}} =
p_{\mathstrut\mbox{\tiny HT}}
= {2 \over 3}$
and that
\(p_{\mathstrut\mbox{\tiny TH}} = {1\over 3}\,\).
Deduce that the probability that A wins the game is \({2\over 3}\,\).
- \(B\) and \(C\) play the game.
Find the probability that \(B\) wins.
- Show that if \(C\) plays \(D\), then \(C\) is more likely to win than \(D\),
but that if \(D\) plays \(A\), then \(D\) is more likely to win than \(A\).
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 \(B\).
- 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 \(3/128\).
- My mother is \(AB\) and I am \(AA\).
Find the probability that my father is \(AB\).
It is known that there are
three manufacturers \(A, B, C,\) who can produce
micro chip MB666. The probability that a randomly selected MB666
is produced by \(A\) is \(2p\), and the corresponding probabilities for
\(B\) and \(C\) are \(p\) and \(1 - 3p\), respectively, where
\({{0} \le p \le {1 \over 3}}.\) It is also known that \(70\%\) of
MB666 micro chips from \(A\) are sound and that the corresponding
percentages for \(B\) and \(C\) are \(80\%\) and \(90\%\), respectively.
Find in terms of \(p\), the conditional probability, \(\P(A {\vert} S)\),
that if a randomly selected
MB666 chip is found to be sound then it came from \(A\), and also the
conditional probability, \(\P(C {\vert} S)\), that
if it is sound then it came from \(C\).
A quality inspector took a random sample
of one MB666 micro chip and found it to be sound. She then traced
its place of manufacture to be \(A\), and
so estimated \(p\) by calculating the value of \(p\)
that corresponds to the greatest value
of \(\P(A {\vert} S)\). A second quality
inspector also a took random sample of one MB666 chip and
found it to be sound. Later he traced its place of manufacture
to be \(C\) and so estimated \(p\) by applying the procedure of his
colleague to \(\P(C {\vert} S)\).
Determine the values of the two estimates and comment
briefly on the results obtained.
I have a bag initially containing \(r\) red fruit pastilles
(my favourites)
and \(b\) fruit pastilles of other colours. From time to time
I shake the bag thoroughly and remove a pastille at random.
(It may be assumed that all pastilles have an equal chance
of being selected.) If the pastille is red I eat it
but otherwise I replace it in the bag. After \(n\) such
drawings, I find that I have only eaten one pastille.
Show that the probability that I ate it on my last drawing
is
\[\frac{(r+b-1)^{n-1}}{(r+b)^{n}-(r+b-1)^{n}}.\]
The diagnostic test AL has a probability 0.9 of giving a positive result when applied to a person suffering from the rare disease mathematitis. It also has a probability 1/11 of giving a false positive result when applied to a non-sufferer. It is known that only \(1\%\) of the population suffer from the disease. Given that the test AL is positive when applied to Frankie, who is chosen at random from the population, what is the probability that Frankie is a sufferer?
In an attempt to identify sufferers more accurately, a second diagnostic test STEP is given to those for whom the test AL gave a positive result. The probablility of STEP giving a positive result on a sufferer is 0.9, and the probability that it gives a false positive result on a non-sufferer is \(p\).
Half of those for whom AL was positive and on whom STEP then also gives a positive result are sufferers. Find \(p\).
Show Solution
\begin{align*}
\mathbb{P}(M | P_{AL}) &= \frac{\mathbb{P}(M \cap P_{AL})}{\mathbb{P}(P_{AL})} \\
&= \frac{\frac{1}{100} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} + \frac{99}{100} \frac{1}{11}} \\
&= \frac{\frac{9}{10}}{\frac{9}{10} + \frac{9}{1}} \\
&= \frac{9}{99} = \frac{1}{11} \\
\end{align*}
\begin{align*}
&& \frac12 &= \mathbb{P}(M | P_{STEP}, P_{AL}) \\
&&&= \frac{\frac{1}{100} \frac{9}{10} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} \frac{9}{10} + \frac{99}{100} \frac{1}{11}p} \\
&&&= \frac{81}{81+900p} \\
\Rightarrow && p &= \frac{81}{900} = \frac{9}{100}
\end{align*}
Therefore \(p = 9\%\)
Write down the probability of obtaining \(k\) heads in \(n\) tosses of a fair coin. Now suppose that \(k\) is known but \(n\) is unknown. A maximum likelihood estimator (MLE) of \(n\) is defined to be a value (which must be an integer) of \(n\) which maximizes the probability of \(k\) heads.
A friend has thrown a fair coin a number of times. She tells you that she has observed one head. Show that in this case there are two MLEs of the number of tosses she has made. She now tells you that in a repeat of the exercise she has observed \(k\) heads. Find the two MLEs of the number of tosses she has made. She next uses a coin biased with probability \(p\) (known) of showing a head, and again tells you that she has observed \(k\) heads. Find the MLEs of the number of tosses made. What is the condition for the MLE to be unique?
Show Solution
\begin{align*}
&& \mathbb{P}(k \text{ heads} | n\text{ tosses}) &= \binom{n}k 2^{-n} \\
&& \mathbb{P}(1 \text{ head} | n\text{ tosses}) &= n2^{-n} \\
\Rightarrow && \frac{ \mathbb{P}(1 \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(1 \text{ head} | n\text{ tosses}) } &= \frac{n+1}{2n}
\end{align*}
Which is less than \(1\) unless \(n \geq 1\). Therefore the MLE is \(n = 1\) or \(n= 2\).
\begin{align*}
\frac{ \mathbb{P}(k \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} | n\text{ tosses}) } &= \frac{\binom{n+1}{k}}{2 \binom{n}{k}} \\
&= \frac{(n+1)!(n-k)!}{2n!(n+1-k)!} \\
&= \frac{n+1}{2(n+1-k)}
\end{align*}
This is less than or equal to \(1\) if \(n+1 = 2(n+1-k) \Leftrightarrow n= 2k-1\), therefore the MLEs are \(2k-1\) and \(2k\).
If the coin is biased, we have
\begin{align*}
&& \frac{ \mathbb{P}(k \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} | n\text{ tosses}) } &= \frac{\binom{n+1}{k}p^kq^{n+1-k}}{\binom{n}{k}p^kq^{n-k}} \\
&&&= \frac{n+1}{(n+1-k)}q \\
\\
&& 1 & \geq \frac{n+1}{(n+1-k)}q \\
\Leftrightarrow && (n+1)(1-q) &\geq k \\
\Leftrightarrow && n+1 & \geq \frac{k}{p}
\end{align*}
Therefore the probability is increasing until \(n+1 \geq \frac{k}{p}\). If \(\frac{k}p\) is an integer the MLEs are \(\frac{k}{p}-1\) and \(\frac{k}p\), otherwise it is \(\lfloor \frac{k}{p} \rfloor\) and the MLE is unique.
Mr Blond returns to his flat to find it in complete darkness. He knows that this means that one of four assassins Mr 1, Mr 2, Mr 3 or Mr 4 has set a trap for him. His trained instinct tells him that the probability that Mr \(i\) has set the trap is \(i/10\). His knowledge of their habits tells him that Mr \(i\) uses a deadly trained silent anaconda with probability \((i+1)/10\), a bomb with probability \(i/10\) and a vicious attack canary with probability \((9-2i)/10\) \([i=1,2,3,4]\).
He now listens carefully and, hearing no singing, concludes correctly that no canary is involved. If he switches on the light and the trap is a bomb he has probability \(1/2\) of being killed
but if the trap is an anaconda he has probability \(2/3\) of survival. If he does not switch on the light and the trap is a bomb he is certain to survive but, if the trap is an anaconda, he has a probability \(1/2\) of being killed. His professional pride means that he must enter the flat.
Advise Mr Blond, giving reasons for your advice.
Show Solution
\begin{array}{c|c|c|c}
& A & B & C \\ \hline
1 & \frac{1}{10} \cdot \frac{2}{10} & \frac{1}{10} \cdot \frac{1}{10} & \frac{1}{10} \cdot \frac{7}{10} \\
2 & \frac{2}{10} \cdot \frac{3}{10} &\frac{2}{10} \cdot \frac{2}{10} &\frac{2}{10} \cdot \frac{5}{10} \\
3 & \frac{3}{10} \cdot \frac{4}{10} &\frac{3}{10} \cdot \frac{3}{10} &\frac{3}{10} \cdot \frac{3}{10} \\
4 & \frac{4}{10} \cdot \frac{5}{10} &\frac{4}{10} \cdot \frac{4}{10} &\frac{4}{10} \cdot \frac{1}{10} \\ \hline
& \frac{2+6+12+20}{100} & \frac{1 + 4 + 9 + 16}{100} & \frac{7 + 10 + 9 + 4}{100}
\end{array}
Therefore \(\mathbb{P}(A) = \frac{4}{10}, \mathbb{P}(B) = \frac{3}{10}, \mathbb{P}(C) = \frac{3}{10}\), in particular,
\begin{align*}
\mathbb{P}(A | \text{not }C) &= \frac{4}{7} \\
\mathbb{P}(B | \text{not }C) &= \frac{3}{7} \\
\end{align*}
If he switches the light on, his probability of survival is \(\frac47 \cdot \frac23 + \frac37 \cdot \frac12 = \frac{25}{42}\), if he doesn't his probability is \(\frac12 \cdot \frac47 +\frac37= \frac{5}{7} = \frac{30}{42}\) therefore he shouldn't switch the light on.
A biased coin, with a probability \(p\) of coming up heads and a probability \(q=1-p\) of coming up tails, is tossed repeatedly. Let \(A\) be the event that the first run of \(r\) successive heads occurs before the first run of \(s\) successive tails. If \(H\) is the even that on the first toss the coin comes up heads and \(T\) is the event that it comes up tails, show that
\begin{alignat*}{1}
\mathrm{P}(A|H) & =p^{\alpha}+(1-p^{\alpha})\mathrm{P}(A|T),\\
\mathrm{P}(A|T) & =(1-q^{\beta})\mathrm{P}(A|H),
\end{alignat*}
where \(\alpha\) and \(\beta\) are to be determined. Use these two equations to find \(\mathrm{P}(A|H),\) \(\mathrm{P}(A|T),\) and hence \(\mathrm{P}(A).\)
Show Solution
\begin{align*}
&& \P(A|H) &= \P(\text{achieve }r\text{ heads immediately}) + \P(\text{don't and then achieve it from having flipped a tail}) \\
&&&= p^{r-1} + (1-p^{r-1}) \cdot \P(A|T) \\
&& \P(A|T) &= (1-q^{s-1})\P(A|H) \\
\\
&&\P(A|H) &= p^{r-1}+(1-p^{r-1})(1-q^{s-1})\P(A|H) \\
\Rightarrow && \P(A|H) &= \frac{p^{r-1}}{1-(1-p^{r-1})(1-q^{s-1})} \\
&& \P(A|T) &= \frac{(1-q^{s-1})p^{r-1}}{1-(1-p^{r-1})(1-q^{s-1})} \\
&& \P(A) &= \frac{(2-q^{s-1})p^{r-1}}{2(1-(1-p^{r-1})(1-q^{s-1}))}
\end{align*}
It has been observed that Professor Ecks proves three types of theorems:
1, those that are correct and new; 2, those that are correct, but
already known; 3, those that are false. It has also been observed
that, if a certain of her theorems is of type \(i\), then her next
theorem is of type \(j\) with probability \(p\low_{ij},\) where \(p\low_{ij}\)
is the entry in the \(i\)th row and \(j\)th column of the following
array:
\[
\begin{pmatrix}0.3 & 0.3 & 0.4\\
0.2 & 0.4 & 0.4\\
0.1 & 0.3 & 0.6
\end{pmatrix}\,.
\]
Let \(a_{i},\) \(i=1,2,3\), be the probability that a given theorem
is of type \(i\), and let \(b_{j}\) be the consequent probability that
the next theorem is of type \(j\).
- Explain why \(b_{j}=a\low_{1}p\low_{1j}+a\low_{2}p\low_{2j}+a\low_{3}p\low_{3j}\,.\)
- Find values of \(a\low_{1},a\low_{2}\) and \(a\low_{3}\) such that \(b_{i}=a_{i}\)
for \(i=1,2,3.\)
- For these values of the \(a_{i}\) find the probabilities \(q\low_{ij}\)
that, if a particular theorem is of type \(j\), then the \textit{preceding
}theorem was of type \(i\).
Bread roll throwing duels at the Drones' Club are governed by a strict
etiquette. The two duellists throw alternatively until one is hit,
when the other is declared the winner. If Percy has probability \(p>0\)
of hitting his target and Rodney has probability \(r>0\) of hitting
his, show that, if Percy throws first, the probability that he beats
Rodney is
\[
\frac{p}{p+r-pr}.
\]
Algernon, Bertie and Cuthbert decide to have a three sided duel in
which they throw in order \(\mathrm{A,B,C,A,B,C,}\ldots\) except that
anyone who is hit must leave the game. Cuthbert always his target,
Bertie hits his target with probability \(3/5\) and Algernon hits his
target with probability \(2/5.\) Bertie and Cuthbert will always aim
at each other if they are both still in the duel. Otherwise they aim
at Algernon. With his first shot Algernon may aim at either Bertie
or Cuthbert or deliberately miss both. Faced with only one opponent
Algernon will aim at him. What are Algernon's changes of winning if
he:
- sep}{3mm}
- \(\bf (i)\) hits Cuthbert with his first shot?
- \(\bf (ii)\) hits Bertie with his first shot?
- \(\bf (iii)\) misses with his first shot?
Advise Algernon as to his best plan and show that, if he uses this
plan, his probability of winning is \(226/475.\)
A message of \(10^{k}\) binary digits is sent along a fibre optic cable
with high probabilities \(p_{0}\) and \(p_{1}\) that the digits 0 and
1, respectively, are received correctly. If the probability of a digit
in the original message being a 1 is \(\alpha,\) find the probability
that the entire message is received correctly.
Find the probability \(\beta\) that a randomly chosen digit in the
message is received as a 1 and show that \(\beta=\alpha\) if, and only
if
\[
\alpha=\frac{q_{0}}{q_{1}+q_{0}},
\]
where \(q_{0}=1-p_{0}\) and \(q_{1}=1-p_{1}.\) If this condition is
satisfied and the received message consists entirely of zeros, what
is the probability that it is correct?
If now \(q_{0}=q_{1}=q\) and \(\alpha=\frac{1}{2},\) find the approximate
value of \(q\) which will ensure that a message of one million binary
digits has a fifty-fifty chance of being received entirely correctly.
The probability of error \(q\) is proportional to the square of the
length of the cable. Initially the length is such that the probability
of a message of one million binary bits, among which 0 and 1 are equally
likely, being received correctly is \(\frac{1}{2}.\) What would this
probability become if a booster station were installed at its mid-point,
assuming that the booster station re-transmits the received version
of the message, and assuming that terms of order \(q^{2}\) may be ignored?
There are 28 colleges in Cambridge, of which two (New Hall and Newnham)
are for women only; the others admit both men and women. Seven women,
Anya, Betty, Celia, Doreen, Emily, Fariza and Georgina, are all applying
to Cambridge. Each has picked three colleges at random to enter on
her application form.
- What is the probability that Anya's first choice college is
single-sex?
- What is the probability that Betty has picked Newnham?
- What is the probability that Celia has picked at least one
single-sex college?
- Doreen's first choice is Newnham. What is the probability that
one of her other two choices is New Hall?
- Emily has picked Newnham. What is the probability that she
has also picked New Hall?
- Fariza's first choice college is single-sex. What is the probability
that she has also chosen the other single-sex college?
- One of Georgina's choices is a single-sex college. What is
the probability that she has also picked the other single-sex college?
Show Solution
- \(\frac{2}{28} = \frac{1}{14}\)
- \(1-\frac{27}{28}\frac{26}{27}\frac{25}{26} = \frac{3}{28}\)
- \(1 - \frac{26}{28}\frac{25}{27}\frac{24}{26} = \frac{13}{63}\)
- \(1-\frac{26}{27}\frac{25}{26} = \frac{2}{27}\)
- \(\frac{1}{27}\)
- There are \(\binom{2}{1} \binom{26}{2} + \binom{2}{2}\binom{26}{1}\) ways to choose at least one single sex college and \( \binom{2}{2}\binom{26}{1}\) ways to choose both, therefore
\begin{align*}
P &= \frac{ \binom{2}{2}\binom{26}{1}}{\binom21 \binom{26}2+ \binom{2}{2}\binom{26}{1}} \\
&= \frac{26}{2\cdot \frac{26\cdot25}{2}+26 }\\
&= \frac{1}{25+1} = \frac{1}{26}
\end{align*}
Calamity Jane sits down to play the game of craps with Buffalo Bill.
In this game she rolls two fair dice. If, on the first throw, the
sum of the dice is \(2,3\) or \(12\) she loses, while if it is \(7\)
or \(11\) she wins. Otherwise Calamity continues to roll the dice until
either the first sum is repeated, in which case she wins, or the sum
is \(7\), in which case she loses. Find the probability that she wins
on the first throw.
Given that she throws more than once, show that the probability that
she wins on the \(n\)th throw is
\[
\frac{1}{48}\left(\frac{3}{4}\right)^{n-2}+\frac{1}{27}\left(\frac{13}{18}\right)^{n-2}+\frac{25}{432}\left(\frac{25}{36}\right)^{n-2}.
\]
Given that she throws more than \(m\) times, where \(m>1,\) what is
the probability that she wins on the \(n\)th throw?
Captain Spalding is on a visit to the idyllic island of Gambriced.
The population of the island consists of the two lost tribes of Frodox
and the latest census shows that \(11/16\) of the population belong
to the Ascii who tell the truth \(3/4\) of the time and \(5/16\) to
the Biscii who always lie. The answers of an Ascii to each question
(even if it is the same as one before) are independent.
Show that the probability that an Ascii gives the same answer twice
in succession to the same question is \(5/8\). Show that the probability
that an Ascii gives the same answer twice is telling the truth is
\(9/10.\)
Captain Spalding addresses one of the natives as follows.
\hspace{1.5em} \textsl{Spalding: }My good man, I'm afraid I'm lost. Should I go left or right to reach the nearest town?\nolinebreak
\hspace{1.5em}\textsl{Native: }Left.
\hspace{1.5em}\textsl{Spalding: }I am a little deaf. Should I go left or right to
reach the nearest town?
\hspace{1.5em}\textsl{Native (patiently): }Left.
Show that, on the basis of this conversation, Captain Spalding should
go left to try and reach the nearest town and that there is a probability
\(99/190\) that this is the correct direction.
The conversation resumes as follows.
\hspace{1.5em}\textsl{Spalding: }I'm sorry I didn't quite hear that. Should I go
left or right to reach the nearest town?
\hspace{1.5em}\textsl{Native (loudly and clearly): }Left.
Shouls Captain Spalding go left or right and why? Show that if he
follows your advice the probability that this is the correct direction
is \(331/628\).
A taxi driver keeps a packet of toffees and a packet of mints in her taxi. From time to time she takes either a toffee (with probability \(p\)) or mint (with probability \(q=1-p\)). At the beginning of the week she has \(n\) toffees and \(m\) mints in the packets. On the \(N\)th
occasion that she reaches for a sweet, she discovers (for the first time) that she has run out of that kind of sweet. What is the probability that she was reaching for a toffee?
Show Solution
\begin{align*}
\mathbb{P}(\text{run out reading for toffee on } N\text{th occassion}) &= \binom{N-1}{n}p^nq^{N-1-n}p
\end{align*}
Since out of the first \(N-1\) times, we need to choose toffee \(n\) times, and then choose it again for the \(N\)th time.
Therefore:
\begin{align*}
\mathbb{P}(\text{reaching for toffee} | \text{run out on }N\text{th occassion}) &= \frac{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion})}{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion}) + \mathbb{P}(\text{reaching for mint and run out on }N\text{th occassion})} \\
&= \frac{ \binom{N-1}{n}p^nq^{N-1-n}p}{ \binom{N-1}{n}p^nq^{N-1-n}p + \binom{N-1}{m}q^mp^{N-1-m}q} \\
&= \frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} \l \frac{q}{p} \r^{m+ n+ 2-N}}
\end{align*}
Some conclusions we can draw from this are:
As \(p \to 1, q \to 0\), the probability they were reaching for a Toffee tends to \(1\). (And vice versa).
If \(p = q\), then the probability is:
\begin{align*}
\frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} }
\end{align*}
Since \(n+1 \leq N \leq n+m+1\) where \(n \geq m\) we can notice that:
\begin{align*}
\text{if } N = m + n + 1 && \binom{m+n+1 - 1}{n} &= \binom{m+n+1 - 1}{m} & \text{ so } \mathbb{P} = \frac12 \\
\text{if } N = n+k && \binom{n+k-1}{n} &< \binom{n+k-1}{m} & \text{ so } \mathbb{P} < \frac12 \\
\end{align*}