Probability

Activity 2.1   Why is $\Omega=\{1,1,2\}$, not a sensible way to try to define a sample space?
$\blacksquare$

Answer 2.1   The proposed definition is not sensible, because the elements of $\Omega$ should all be different, but there are two elements 1 in this proposed $\Omega$.
$\blacksquare$

Activity 2.2   Write out all the events for the sample space $\Omega=\{a,b,c\}$. (There are eight of them.)
$\blacksquare$

Answer 2.2   The events are $\emptyset$, $\{a\}$, $\{b\}$, $\{c\}$, $\{a,b\}$, $\{b,c\}$, $\{a,c\}$, $\Omega$.
$\blacksquare$

Activity 2.3   Prove, using only the axioms of probability given, that $P(\emptyset)=0$. (Hint: $\emptyset\cup \emptyset =\emptyset$ and $\emptyset\cap \emptyset =\emptyset$.)
$\blacksquare$

Answer 2.3   Since $\emptyset\cap \emptyset =\emptyset$, $\emptyset$ is disjoint from $\emptyset$, we can apply the addition rule to get

$$P(\emptyset\cup \emptyset) = P(\emptyset) +P(\emptyset),$$

which is

$$P(\emptyset) = 2P(\emptyset),$$

It follows by simple algebra that $P(\emptyset)=0$.
$\blacksquare$

Activity 2.4   If all elementary outcomes are equally likely, and $\Omega=\{a,b,c,d\}$, $A=\{a,b,c\}$, $B=\{c,d\}$, find $P(A\vert B)$ and $P(B\vert A)$.
$\blacksquare$

Answer 2.4   $\Omega$ has 4 elementary outcomes that are equally likely, so each elementary outcome has probability 1/4.

$$P(A\vert B) =P(A\cap B)/P(B)=P(\{c\})/P(\{c,d\})$$

$$=\frac{1/4}{1/4+1/4}=1/2.$$

$$P(B\vert A) =P(B\cap A)/P(A)=P(\{c\})/P(\{a,b,c\})$$

$$=\frac{1/4}{1/4+1/4+1/4}=1/3.$$

$\blacksquare$

Activity 2.5   Identify the events and then the conditional probabilities wrongly equated in the prosecutors fallacy above.
$\blacksquare$

Answer 2.5   This is also an example of how slippery probability statements can be when the outcome space is not well defined, and when they are stated without care in ordinary language. How many readers of this material thought that the prosecutor's argument was correct? And yet it is an argument wholly lacking in coherence.

Let's imagine that for his first statement the prosecutor is thinking about a large population of people similar to the accused, say all men in the UK. The prosecutor's first statement, based on the knowledge of the genetics of that population, is that for a person taken at random from that population, the conditional probability of a DNA match given that the person was not at the crime scene is 1 in a million. We could write

$$P($$DNA match$$\vert$$Accused not at crime scene$$)=1/1000000.$$

The prosecutor's next statement is that the probability of the accused being at the crime scene given that there is a DNA match is 1000000/1000001. We could write

$$P($$Accused at crime scene$$\vert$$DNA match$$)= 1000000/1000001,$$

which is

$$P($$Accused not at crime scene$$\vert$$DNA match$$)= 1/1000001.$$

Its is clear that the prosecutor wants the jury to believe that $P(A\vert B)=P(B\vert A)$ for event $A =$DNA match and event $B=$Accused not at crime scene. However, if $P(A)$ and $P(B)$ are different from each other, this equality will not be true. The prosecutor wishes, in effect, that the jury treats $A$ and $B$ as the same event, but they are not the same.
$\blacksquare$

Activity 2.6   Suppose that we toss a coin twice. The sample space is $\Omega=\{HH,HT,TH,TT\}$, where the elementary outcomes are defined in the obvious way - for instance $HT$ is heads on the first toss and tails on the second toss. Show that if all four elementary outcomes are equally likely, then the events `Heads on the first toss' and `Heads on the second toss' are independent.
$\blacksquare$

Answer 2.6   Note carefully here that we just assume equally likely elementary outcomes, so that each has probability 1/4, and the independence follows.

The event `Heads on the first toss' is $A=\{HH, HT\}$ and has probability 1/2, for it is specified by two elementary outcomes. The event `Heads on the second toss' is $B=\{HH, TH\}$ and has probability 1/2. The event `Heads on both the first and the second toss' is $A\cap B=\{HH\}$ and has probability 1/4. So the multiplication property $P(A\cap B)=1/4=1/2\times 1/2=P(A)P(B)$ is satisfied, and the two events are independent.
$\blacksquare$

Activity 2.7   Show that if $A$ and $B$ are disjoint events, and are also independent, then $P(A)=0$ or $P(B)=0$.
$\blacksquare$

Answer 2.7   It's important to get the logical flow in the right direction here. We are told that $A$ and $B$ are disjoint events, that is

$$A\cap B=\emptyset.$$

So

$$P(A\cap B)=0.$$

We are also told that $A$ and $B$ are independent, that is

$$P(A\cap B)=P(A)P(B).$$

It follows that

$$0=P(A)P(B)$$

and so either $P(A)=0$ or $P(B)=0$.
$\blacksquare$

Activity 2.8   Prove this result from first principles.
$\blacksquare$

Answer 2.8   From the definition of conditional probability we know

$$P(B_i\vert A)=\frac{P(B_i\cap A)}{P(A)},$$

and that

$$P(B_i\cap A)=P(A\cap B_i)=P(A\vert B_i)P(B_i).$$

It remains to show that

$$P(A)=\sum_{j=1}^n P(A\vert B_j)P(B_j).$$

Now,

$$A=A\cap \Omega=A\cap \bigcup_{j=1}^n B_j$$

and using a distributive law for sets

$$A\cap \bigcup_{j=1}^n B_j=\bigcup_{j=1}^n (A\cap B_j).$$

Since the $B_j$ are disjoint events and $A\cap B_j$ is included in $B_j$, the events $A\cap B_j$ are disjoint. We can use the addition rule for probabilities to get

$$P(A)=P(\bigcup_{j=1}^n (A\cap B_j))=\sum_{j=1}^n P(A\cap B_j).$$

It only remains to notice, as we did before, that from the definition of conditional probability, for every $j$

$$P(A\cap B_j)=P(A\vert B_j)P(B_j).$$

$\blacksquare$

Activity 2.9   Plagiarism is a serious problem for assessors of course-work. One check on plagiarism is to compare the course-work with a standard text. If the course-work has plagiarised that text, then there will be a 95% chance of finding exactly two phrases that are the same in both course-work and text, and a 5% chance of finding three or more. If the work is not plagiarised, then these probabilities are both 50%.

Suppose that 5% of course-work is plagiarised. An assessor chooses some course-work at random. What is the probability that it has been plagiarised if it has exactly two phrases in the text? And if there are three phrases? Did you manage to get a roughly correct guess of these results before calculating?
$\blacksquare$

Answer 2.9   Suppose that two phrases are the same. We use Bayes' theorem.

$$P($$plagiarised$$\vert$$two the same$$)=\frac{0.95\times 0.05}{0.95\times 0.05+0.5\times 0.95}=0.0909.$$

Finding two phrases has increased the chance the work is plagiarised from 5% to 9.1%. Did you get anywhere near 9% when guessing? Now suppose that we find three or more phrases.

$$P($$plagiarised$$\vert$$three the same$$)=\frac{0.05\times 0.05}{0.05\times 0.05+0.5\times 0.95}=0.0052.$$

No plagiariser is silly enough to make three or more phrases the same, so if we find three or more, the chance of the work being plagiarised falls from 5% to 0.5%. How close did you get by guessing?
$\blacksquare$

Activity 2.10   A box contains three red balls and two green balls. Two balls are taken from it without replacement. What is the probability that 0 balls taken are red? And 1 ball? And 2 balls? Show that the probability that the first ball taken is red is the same as the probability that the second ball taken is red.
$\blacksquare$

Answer 2.10   We have, in the notation used for sampling without replacement, $n=2$, $N=5$, $R=3$. The formula is

$$P(r$$    red balls$$)=\frac{\binom{R}{r}\binom{N-R}{n-r}}{\binom{N}{n}}$$

and so

$$P(0$$    red balls$$)=\frac{\binom{3}{0}\binom{2}{2}}{\binom{5}{2}}=\frac{1\times 1}{10}=\frac{1}{10},$$

$$P(1$$    red balls$$)=\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}}=\frac{3\times 2}{10}=6/10,$$

$$P(2$$    red balls$$)=\frac{\binom{3}{2}\binom{2}{0}}{\binom{5}{2}}=\frac{3\times 1}{10}=3/10.$$

Note that these three probabilities must add to 1.

If the balls are labelled from 1 to 5, every possible permutation of two of the numbers from 1 to 5 is an equally likely way in which we can draw the first and second balls. Since each number is equally often in first and second place in these permutations, the number of red balls in first draws must be the same as the number of red balls on second draws.
$\blacksquare$