At any instant the probability that it is safe to cross a busy road is \(0.1\). A toad is waiting to cross this road. Every minute she looks at the road. If it is safe, she will cross; if it is not safe, she will wait for a minute before attempting to cross again. Find the probability that she eventually crosses the road without mishap.
Later on, a frog is also trying to cross the same road. He also inspects the traffic at one minute intervals and crosses if it is safe. Being more impatient than the toad, he may also attempt to cross when it is not safe. The probability that he will attempt to cross when it
is not safe is \(n/3\) if \(n\leqslant3,\) where \(n\) minutes have elapsed since he firrst inspected the road. If he attempts to cross when it is not safe, he is run over with probability \(0.8,\) but otherwise he reaches the other side safely. Find the probability that he eventually crosses the road without mishap.
What is the probability that both reptiles safely cross the road with the frog taking less time than the toad? If the frog has not arrived at the other side 2 minutes after he began his attempt to cross, what is the probability that the frog is run over (at some stage) in his attempt to cross?
\textit{[Once moving, the reptiles spend a negligible time on their attempt
to cross the road.]}
Show Solution
Since the toad never crosses when it's not safe, she is certain to cross. (Probability she hasn't crossed after the \(n\)th minute is \(0.9^n \to 0\)).
\begin{array}{c|c|c|c|c|c|c|c}
\text{will try dangerously} & \text{is safe} & \text{has tried} & \text{tries safely} & \text{tries unsafely} & \text{succeeds} & \text{succeeds unsafely} & \text{fails} \\ \hline
0 & 0.1 & 0 & 0.1 & 0 & 0.1 & 0 & 0\\
\frac13 & 0.1 & 0.1 & 0.09 & 0.27 & 0.144 & 0.054 & 0.216\\
\frac23 & 0.1 & 0.46 & 0.054 & 0.324 & 0.1188 & 0.0648 & 0.2592\\
1 & 0.1 & 0.838 & 0.0162 & 0.1458 & 0.04536 & 0.02916 & 0.11664\\ \hline
& & & & & 0.40816 & 0.14796 & \\
\hline
\end{array}
So \(\mathbb{P}(\text{frog crosses safely}) = 0.40816\) and \(\mathbb{P}(\text{frog beats toad across}) = 0.14796\).
\begin{align*}
\mathbb{P}(\text{frog run over} | \text{frog not crossed after 2 minutes}) &= \frac{\mathbb{P}(\text{frog run over and frog not crossed after 2 minutes})}{\mathbb{P}(\text{frog not crossed after 2 minutes})} \\
&= \frac{\mathbb{P}(\text{frog run over within 2 minutes})}{\mathbb{P}(\text{frog not crossed after 2 minutes})} \\
&= \frac{\mathbb{P}(\text{frog run over within 2 minutes})}{1-\mathbb{P}(\text{crossed after 2 minutes})} \\
&= \frac{0.216+0.2592}{1-0.3628} \\
&= 0.7457\ldots
\end{align*}
A bus is supposed to stop outside my house every hour on the hour. From long observation I know that a bus will always arrive some time between 10 minutes before and ten minutes after the hour. The probability it arrives at a given instant increases linearly (from zero at 10 minutes before the hour) up to a maximum value at the hour, and then decreases linearly at the same rate after the hour. Obtain the probability density function of \(T\), the time in minutes after the scheduled time at which a bus arrives.
If I get up when my alarm clock goes off, I arrive at the bus stop at 7.55am. However, with probability 0.5, I doze for 3 minutes before it rings again. In that case with probability 0.8 I get up then and reach the bus stop at 7.58am, or, with probability 0.2, I sleep a little longer, not reaching the stop until 8.02am. What is the probability that I catch a bus by 8.10am?
I buy a louder alarm clock which ensures that I reach the stop at exactly the same time each morning. This clock keeps perfect time, but may be set to an incorrect time. If it is correct, the alarm goes off so that I should reach the stop at 7.55am. After 100 mornings I find that I have had to wait for a bus until after 9am (according to the new clock) on 5 occasions. Is this evidence that the new clock is incorrectly set?
{[}The time of arrival of different buses are independent of each
other.{]}
Show Solution
The probability density function will look like a triangle with base \(20\) minutes and therefore height \(\frac{1}{10}\) per minute, ie:
\begin{align*}
f_T(t) &= \begin{cases} \frac{1}{100}(t+10) & \text{if } -10 \leq t \leq 0 \\ \frac{1}{100}(10-t) & \text{if } 0 \leq t \leq 10 \\ 0 & \text{otherwise} \end{cases}
\end{align*}
\begin{align*}
\mathbb{P}(\text{catch bus}) &=0.5 \mathbb{P}(\text{bus arrives after 7:55})+0.4 \mathbb{P}(\text{bus arrives after 7:58}) + 0.1 \mathbb{P}(\text{bus arrives after 8:02}) \\
&= \frac12 \cdot \left (1 - \frac18 \right) + \frac{2}{5} \cdot \left ( 1 - \frac{4^2}{5^2} \cdot \frac{1}{2} \right) + \frac{1}{10} \cdot \frac{4^2}{5^2} \cdot \frac12 \\
&= \frac{1\,483}{2\,000} \\
&\approx 74\%
\end{align*}
\begin{align*}
\mathbb{P}(\text{catch bus}) &= \mathbb{P}(\text{bus arrives after 7:55}) \mathbb{P}(\text{catch next bus by 9:00}) \\
&= \frac78 + \frac18 \cdot \frac12 \\
&= \frac{15}{16}
\end{align*}
He should expect to miss \(6.25\) buses, so missing \(5\) seems about right. (Using a binomial calculation, seeing 5 or fewer buses is ~\(40\%\) which isn't suspicious).
Each day, I choose at random between my brown trousers, my grey trousers
and my expensive but fashionable designer jeans. Also in my wardrobe,
I have a black silk tie, a rather smart brown and fawn polka-dot tie,
my regimental tie, and an elegant powder-blue cravat which I was given
for Christmas. With my brown or grey trousers, I choose ties (including
the cravat) at random, except of course that I don\textquoteright t
wear the cravat with the brown trousers or the polka-dot tie with
the grey trousers. With the jeans, the choice depends on whether it
is Sunday or one of the six weekdays: on weekdays, half the time I
wear a cream-coloured sweat-shirt with \(E=mc{}^{2}\) on the front
and no tie; otherwise, and on Sundays (when naturally I always wear
a tie), I just pick at random from my four ties.
This morning, I received through the post a compromising photograph
of myself. I often receive such photographs and they are equally likely
to have been taken on any day of the week. However, in this particular
photograph, I am wearing my black silk tie. Show that, on the basis
of this information, the probability that the photograph was taken
on Sunday is \(11/68\).
I should have mentioned that on Mondays I lecture on calculus and
I therefore always wear my jeans (to make the lectures seem easier
to understand). Find, on the basis of the complete information, the
probability that the photograph was taken on Sunday.
[The phrase `at random' means `with equal probability'.]
A and B play a guessing game. Each simultaneously names one of the numbers \(1,2,3.\) If the numbers differ by 2, whoever guessed the smaller pays the opponent £\(2\). If the numbers differ by 1, whoever guessed the larger pays the opponent £\(1.\)
Otherwise no money changes hands. Many rounds of the game are played.
- If A says he will always guess the same number \(N\), explain (for each value of \(N\)) how B can maximise his winnings.
- In an attempt to improve his play, A announces that he will guess each number at random with probability \(\frac{1}{3},\) guesses on different rounds being independent. To counter this, B secretly decides
to guess \(j\) with probability \(b_{j}\) (\(j=1,2,3,\, b_{1}+b_{2}+b_{3}=1\)), guesses on different rounds being independent. Derive an expression for B's expected winnings on any round. How should the probabilities \(b_{j}\) be chosen so as to maximize this expression?
- A now announces that he will guess \(j\) with probability \(a_{j}\) (\(j=1,2,3,\, a_{1}+a_{2}+a_{3}=1\)). If B guesses \(j\) with probability \(b_{j}\) (\(j=1,2,3,\, b_{1}+b_{2}+b_{3}=1\)), obtain an expression for his expected winnings in the form
\[
Xa_{1}+Ya_{2}+Za_{3}.
\]
Show that he can choose \(b_{1},b_{2}\) and \(b_{3}\) such that \(X,Y\) and \(Z\) are all non-negative. Deduce that, whatever values for \(a_{j}\) are chosen by A, B can ensure that in the long run he loses no money.
Show Solution
- Suppose A always plays \(1\), then B should always play \(2\) and every time they will win 1.
Suppose A always plays \(2\) then B should always play \(3\) and every time they will win 1.
If A always plays \(3\) then B should always play \(1\) and every time they will win 2.
- \begin{array}{cccc}
& b_1 & b_2 & b_3 \\
\frac13 & (0, \frac{b_1}{3}) & (1, \frac{b_2}{3}) & (-2, \frac{b_3}{3}) \\
\frac13 & (-1, \frac{b_1}{3}) & (0, \frac{b_2}{3}) & (1, \frac{b_3}{3}) \\
\frac13 & (2, \frac{b_1}{3}) & (-1, \frac{b_2}{3}) & (0, \frac{b_3}{3}) \\
\end{array}
Therefore the expected value is: \(\frac{b_1}{3} - \frac{b_3}{3}\) and to maximise this he should always guess \(1\) (ie \(b_1 = 1, b_2 = 0, b_3 = 0\).)
- \begin{array}{cccc}
& b_1 & b_2 & b_3 \\
a_1 & (0, a_1b_1) & (1, a_1b_2) & (-2, a_1b_3) \\
a_2 & (-1, a_2b_1) & (0, a_2b_2) & (1, a_2b_3) \\
a_3 & (2, a_3b_1) & (-1, a_3b_2) & (0, a_3b_3) \\
\end{array}
Therefore the expected value is:
\((b_2-2b_3)a_1 + (b_3-b_1)a_2 + (2b_1-b_2)a_3\)
We need \(b_2 \geq 2b_3, b_3 \geq b_1, 2b_1 \geq b_2\) so \(b_1 \leq b_3 \leq \frac12 b_2 \leq b_1\) so we could take \(b_1 = b_3 = \frac12 b_2\) or \(b_1 = b_3 = \frac14, b_2 = \frac12\) and all values would be \(0\).
Therefore by choosing these values \(B\) can guarantee his expected value is \(0\) and therefore shouldn't expect to lose money in the long run.
My two friends, who shall remain nameless, but whom I shall refer to as \(P\) and \(Q\), both told me this afternoon that there is a body in my fridge. I'm not sure what to make of this, because \(P\) tells the truth with a probability of only \(p\), while \(Q\) (independently) tells the truth with probability \(q\). I haven't looked in the fridge for some time, so if you had asked me this morning, I would have said
that there was just as likely to be a body in it as not. Clearly, in view of what \(P\) and \(Q\) told me, I must revise this estimate.
Explain carefully why my new estimate of the probability of there being a body in the fridge should be
\[
\frac{pq}{1-p-q+2pq}.
\]
I have now been to look in the fridge, and there is indeed a body in it; perhaps more than one. It seems to me that only my enemy \(A\), or my enemy \(B\), or (with a bit of luck) both \(A\) and \(B\) could be in my fridge, and this morning I would have judged these three possibilities to be equally likely. But tonight I asked \(P\) and \(Q\) separately whether or not \(A\) was in the fridge, and they each said that he was. What should be my new estimate of the probability that both \(A\) and \(B\) are in my fridge?
Of course, I tell the truth always.
Show Solution
\begin{align*}
\mathbb{P}(\text{body in fridge} | \text{P and Q say so}) &= \frac{\mathbb{P}(\text{body in fridge and P and Q say so})}{\mathbb{P}(\text{P and Q say so})} \\
&= \frac{\frac12 pq}{\mathbb{P}(\text{body in fridge and P and Q say so})+\mathbb{P}(\text{no body in fridge and P and Q say so})} \\
&= \frac{\frac12 pq}{\frac12 pq + \frac12(1-p)(1-q)} \\
&= \frac{pq}{pq + 1-p-q+pq} \\
&= \frac{pq}{1-p-q+2pq}
\end{align*}
\begin{align*}
\mathbb{P}(\text{A and B in fridge} | \text{P and Q say A is in fridge}) &= \frac{\mathbb{P}(\text{A and B in fridge and P and Q say A is in fridge}) }{\mathbb{P}( \text{P and Q say A is in fridge}) } \\
&= \frac{\frac13pq}{\frac13pq+\frac13pq+\frac13(1-p)(1-q)} \\
&= \frac{pq}{1-p-q+3pq}
\end{align*}
A patient arrives with blue thumbs at the doctor's surgery. With probability \(p\) the patient is suffering from Fenland fever and requires treatment costing \(\pounds 100.\) With probability \(1-p\) he is suffering from Steppe syndrome and will get better anyway. A test exists which infallibly gives positive results if the patient is suffering from Fenland fever but also has probability \(q\) of giving positive results if the patient is not. The test cost \(\pounds 10.\) The doctor decides to proceed as follows. She will give the test repeatedly until either the last test is negative, in which case she dismisses the patient with kind words, or she has given the test \(n\) times with positive results each time, in which case she gives the treatment. In the case \(n=0,\) she treats the patient at once. She wishes to minimise the expected cost \(\pounds E_{n}\) to the National Health Service.
- Show that
\[
E_{n+1}-E_{n}=10p-10(1-p)q^{n}(9-10q),
\]
and deduce that if \(p=10^{-4},q=10^{-2},\) she should choose \(n=3.\)
- Show that if \(q\) is larger than some fixed value \(q_{0},\) to be determined explicitly, then whatever the value of \(p,\) she should choose \(n=0.\)
Show Solution
- \(E_{n+1} - E_n\) is the additional cost of the extra test \(10p+10(1-p)q^n\) from people who have yet to fail a test plus the reduce cost of people who will fail the final test, \(-100(1-p)q^n(1-q)\)
\begin{align*}
E_{n+1}-E_{n} &= 10p+10(1-p)q^n-100(1-p)q^n(1-q) \\
&=10p +10(1-p)q^n(1-10(1-q)) \\
&= 10p +10(1-p)q^n(-9+10q) \\
&= 10p - 10(1-p)q^n(9-10q)
\end{align*}
\begin{align*}
&& 10p - 10(1-p)q^n(9-10q) &> 0 \\
\Leftrightarrow && \frac{p}{(1-p)(9-10q)} &>q^n
\end{align*}
If \(p = 10^{-4}, q = 10^{-2}\) we have:
\begin{align*}
\frac{p}{(1-p)(9-10q)} &= \frac{10^{-4}}{(1-10^{-4})(9-10^{-3})} \\
&\approx 10^{-5}
\end{align*}
and \(q^2 < 10^{-5} < q^3\) so she should stop after the 3rd test.
- She shouldn't bother testing if \begin{align*}
&& \frac{p}{(1-p)(9-10q)} &>1 \\
\Leftrightarrow && \frac{p}{1-p} &>9-10q \\
\Leftrightarrow && 10q &>9 \\
\Leftrightarrow && q &> \frac9{10} = q_0
\end{align*}