Most of you will have now done Examples Sheet 2. Let me make a few comments about some things I hope you learn from that sheet. Some of what I say below is "in brief', and you may need to fill in details or ask your supervisor.
#3,4. There is often more than one nice way to solve a problem. I think that both of these problems have at least two nice methods of solution. Have you been able to think of more than one way to solve them? For example, #4 can be solved by summing a geometric series. It can also be solved via a simple order 0 recurrence relation.
#6 (ii). You should have discovered that $Y$ and $X$ are independent. So $X=10$ provides no information about $Y$. What is the intuition for this? Well, we might imagine that $X$ and $Y$ are initially numbers of "detectable" and "non-detectable" misprints, created as independent samples from the Poission distribution with parameter $\lambda/2$. The page has $X+Y$ misprints (which is distributed Poisson, parameter $\lambda$). The proofreader finds the detectable ones. Clearly, in doing this she discovers nothing about $Y$. This model gives the same distribution of $X$ and $Y$ as one in which there are $X+Y$ misprints and each is detected with probability $1/2$.
#7,8. The results in these questions are often proved by students in something like 2 pages of working. But they can also be set out in perhaps only 2-4 lines of working. Make sure you learn (from your supervisor) the fast way to derive these results.
#10. There are several ways to show that $Y_1,\dotsc,Y_n$ are independent. One way is via the fact that $P(Y_1=1,Y_2=1,\dotsc,Y_n=1)=1/n!=P(Y_1=1)\cdots P(Y_n=1)$. Then apply the result of question #2.
#12. The coupon collector's problem. This question is one that should lodge permanently in your brain. There are many problems in probability, design of computer science algorithms, and elsewhere, which at some point involve this problem. I think it is interesting that the expected time to collect the $n$ figures is about $n\log n$. However, the expected time needed to collect $99\%$ of the figures is only about $n\log 100$. So it is the last fraction of figures that are costly to collect.
#14. Did you notice that this method would work for simulating any probability $\alpha\in (0,1)$? For example, if we want $P(A)=1/5$, we write this in binary as $\alpha=0.0011001100...$ and then toss a coin. If we see TTHHH we would say $A$ has not occurred, because translating this to $.00111$, it differs from the binary form of $\alpha$ by representing a number $>\alpha$.
#16. This questions has been set to IA students for many years. In my experience few students are able to solve it. But the solution is so beautiful that it would be a pity if your supervisor were not given the opportunity to explain it to you.
#17,18. When I say, "This section is for enthusiasts", I am hoping that most students will think "This means me". After all, if you are reading Mathematics at Cambridge you are by definition a mathematics enthusiast. Even if you cannot solve these questions fully you should at least make some attempt. I would expect almost all students to be able to do #17 for the case $k=2$, $n=3$. It is a good general principle to tackle a difficult problem by first trying to solve a simple special case.
#3,4. There is often more than one nice way to solve a problem. I think that both of these problems have at least two nice methods of solution. Have you been able to think of more than one way to solve them? For example, #4 can be solved by summing a geometric series. It can also be solved via a simple order 0 recurrence relation.
#6 (ii). You should have discovered that $Y$ and $X$ are independent. So $X=10$ provides no information about $Y$. What is the intuition for this? Well, we might imagine that $X$ and $Y$ are initially numbers of "detectable" and "non-detectable" misprints, created as independent samples from the Poission distribution with parameter $\lambda/2$. The page has $X+Y$ misprints (which is distributed Poisson, parameter $\lambda$). The proofreader finds the detectable ones. Clearly, in doing this she discovers nothing about $Y$. This model gives the same distribution of $X$ and $Y$ as one in which there are $X+Y$ misprints and each is detected with probability $1/2$.
#7,8. The results in these questions are often proved by students in something like 2 pages of working. But they can also be set out in perhaps only 2-4 lines of working. Make sure you learn (from your supervisor) the fast way to derive these results.
#10. There are several ways to show that $Y_1,\dotsc,Y_n$ are independent. One way is via the fact that $P(Y_1=1,Y_2=1,\dotsc,Y_n=1)=1/n!=P(Y_1=1)\cdots P(Y_n=1)$. Then apply the result of question #2.
#12. The coupon collector's problem. This question is one that should lodge permanently in your brain. There are many problems in probability, design of computer science algorithms, and elsewhere, which at some point involve this problem. I think it is interesting that the expected time to collect the $n$ figures is about $n\log n$. However, the expected time needed to collect $99\%$ of the figures is only about $n\log 100$. So it is the last fraction of figures that are costly to collect.
#14. Did you notice that this method would work for simulating any probability $\alpha\in (0,1)$? For example, if we want $P(A)=1/5$, we write this in binary as $\alpha=0.0011001100...$ and then toss a coin. If we see TTHHH we would say $A$ has not occurred, because translating this to $.00111$, it differs from the binary form of $\alpha$ by representing a number $>\alpha$.
#16. This questions has been set to IA students for many years. In my experience few students are able to solve it. But the solution is so beautiful that it would be a pity if your supervisor were not given the opportunity to explain it to you.
#17,18. When I say, "This section is for enthusiasts", I am hoping that most students will think "This means me". After all, if you are reading Mathematics at Cambridge you are by definition a mathematics enthusiast. Even if you cannot solve these questions fully you should at least make some attempt. I would expect almost all students to be able to do #17 for the case $k=2$, $n=3$. It is a good general principle to tackle a difficult problem by first trying to solve a simple special case.
