W02 - Examples

Bayes’ Theorem

Bayes’ Theorem: COVID tests

Bayes’ Theorem: COVID tests

Assume that 0.5% of people have COVID. Suppose a COVID test gives a (true) positive on 96% of patients who have COVID, but gives a (false) positive on 2% of patients who do not have COVID. Bob tests positive. What is the probability that Bob has COVID?

Solution

(1) Label events:

• Event $A_P$: Bob is actually positive for COVID
• Event $A_N$: Bob is actually negative; note $A_N=A_P^c$
• Event $T_P$: Bob tests positive
• Event $T_N$: Bob tests negative; note $T_N=T_P^c$

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(2) Identify known data:

• Know: $P[T_P\mid A_P]=96\%$
• Know: $P[T_P\mid A_N]=2\%$
• Know: $P[A_P]=0.5\%$ and therefore $P[A_N]=99.5\%$

We seek: $P[A_P\mid T_P]$

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(3) Translate Bayes’ Theorem:

Set $A=T_P$ and $B=A_P$ in Bayes’:

$$P[A_P\mid T_P]=P[A_P]\cdot \frac{P[T_P\mid A_P]}{P[T_P]}$$

We know all values on the right except $P[T_P]$

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(4) Denominator: apply Total Probability (Division into Cases):

Observe that $T_P\cap A_P$ and $T_P\cap A_N$ are exclusive events, and that:

$$T_P=T_P\cap A_P\bigcup T_P\cap A_N.$$

Therefore:

$$P[T_P]=P[A_P]\cdot P[T_P\mid A_P]+P[A_N]\cdot P[T_P\mid A_N]$$

Plug in data and compute:

$$\gg \gg P[T_P]=\frac{5}{1000}\cdot \frac{96}{100}+\frac{995}{1000}\cdot \frac{2}{100}\gg \gg \approx 0.0247$$

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(5) Plug in and compute:

$$\begin{array}{c}P[A_P\mid T_P]=P[A_P]\cdot \frac{P[T_P\mid A_P]}{P[T_P]}\\ \\ \gg \gg 0.96\cdot \frac{0.005}{0.0247}\gg \gg \approx 19\%\end{array}$$

Intuition - COVID testing

Some people find this low number surprising. In order to repair your intuition, think about it like this: roughly 2.5% of tests are positive, with roughly 2% coming from false positives, and roughly 0.5% from true positives. Only $1/5$ of all the positive results are true ones!

(This rough approximation assumes that $96\%\approx 100\%$.)

If two tests both come back positive, the odds of COVID are now 98%.

If only people with symptoms are tested, so that, say, 20% of those tested have COVID, that is, $P[A_P\mid T_P]=20\%$, then one positive test implies a COVID probability of 92%.

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Inferring bin from marble

Inferring bin from marble

There are marbles in bins in a room:

• Bin 1 holds 7 red and 5 green marbles.
• Bin 2 holds 4 red and 3 green marbles.

Your friend goes in the room, shuts the door, and selects a random bin, then draws a random marble. (Equal odds for each bin, then equal odds for each marble in that bin.) He comes out and shows you a red marble.

What is the probability that this red marble was taken from Bin 1?

Solution

(1) Label events:

• Event $B_1$: friend chooses Bin 1.
• Event $B_2$: friend chooses Bin 2.
• Event $R$: friend draws a red marble.
• Event $G$: friend draws a green marble.

Answer will be $P[B_1\mid R]$.

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(2) Identify knowns:

• Know $P[R\mid B_1]=7/12$.
• Know $P[G\mid B_1]=5/12$.
• Know $P[R\mid B_2]=4/7$.
• Know $P[G\mid B_2]=3/7$.
• Know $P[B_1]=P[B_2]=1/2$.

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(3) Apply Bayes’ Theorem for $P[B_1\mid R]$::

$$P[B_1\mid R]=P[R\mid B_1]\cdot \frac{P[B_1]}{P[R]}$$

Division into Cases for the denominator:

$$P[R]=P[B_1]\cdot P[R\mid B_1]+P[B_2]\cdot P[R\mid B_2]$$

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(3) Plug in data and compute:

$$\begin{array}{c}P[R]\gg \gg \frac{1}{2}\cdot \frac{7}{12}+\frac{1}{2}\cdot \frac{4}{7}\gg \gg \frac{97}{168}\\ \\ P[B_1\mid R]\gg \gg \frac{7}{12}\cdot \frac{1/2}{97/168}\gg \gg \frac{49}{97}\end{array}$$

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Independence

Independence and complements

Independence and complements

Prove that these are logically equivalent statements:

• $A$ and $B$ are independent
• $A$ and $B^c$ are independent
• $A^c$ and $B^c$ are independent

Make sure you demonstrate both directions of each equivalency.

Solution

(1) Assume $P[AB]=P[A]\cdot P[B]$ and show $P[A{B}^c]=P[A]\cdot P[B^c]$:

Divide $A$ into the $B$ cases:

$$P[A]=P[AB]+P[A{B}^c]$$

Apply the assumption:

$$\gg \gg P[A]=P[A]\cdot P[B]+P[A{B}^c]$$

Algebra:

$$\gg \gg P[A](1-P[B])=P[A{B}^c]$$

Negation rule:

$$\gg \gg P[A]\cdot P[B^c]=P[A{B}^c]$$

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(2) Assume $P[A{B}^c]=P[A]\cdot P[B^c]$ and show $P[AB]=P[A]\cdot P[B]$:

Reason the same way as above, but apply the new assumption to break up the second term instead of the first.

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(3) Show that $P[A{B}^c]=P[A]\cdot P[B^c]$ and $P[A^c{B}^c]=P[A^c]\cdot P[B^c]$ are equivalent:

To do this, simply notice that in the first equivalence we can replace $A$ with $B^c$ and $B$ with $A$. Use $AB=BA$ too.

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Independence by hand: red and green marbles

Independence by hand: red and green marbles

A bin contains 4 red and 7 green marbles. Two marbles are drawn.

Let $R_1$ be the event that the first marble is red, and let $G_2$ be the event that the second marble is green.

(a) Show that $R_1$ and $G_2$ are independent if the marbles are drawn with replacement.

(b) Show that $R_1$ and $G_2$ are not independent if the marbles are drawn without replacement.

Solution

(a) With replacement.

Identify knowns:

• Know: $P[R_1]=\frac{4}{11}$
• Know: $P[G_2]=\frac{7}{11}$

Now compute both sides of independence relation:

$$P[R_1{G}_2]=P[R_1]\cdot P[G_2]$$

The right side is $\frac{4}{11}\cdot \frac{7}{11}$.

For $P[R_1{G}_2]$, we have $4\cdot 7$ ways to get $R_1{G}_2$, and $11^2$ total outcomes. So left side is $\frac{4\cdot 7}{11^2}$, which equals the right side.

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(b) Without replacement. This is a bit harder.

(1) Identify knowns:

Know: $P[R_1]=\frac{4}{11}$ and therefore $P[R_1^c]=\frac{7}{11}$

We seek: $P[G_2]$ and $P[R_1{G}_2]$

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(2) Find $P[G_2]$ using Total Probability (Division into Cases):

$$\begin{array}{c}{G}_2=G_2\cap R_1\bigcup G_2\cap R_1^c\\ \\ \gg \gg P[G_2]=P[R_1]\cdot P[G_2\mid R_1]+P[R_1^c]\cdot P[G_2\mid R_1^c]\end{array}$$

Find RHS factors by counting, then compute:

$$\gg \gg P[G_2]=\frac{4}{11}\cdot \frac{7}{10}+\frac{7}{11}\cdot \frac{6}{10}\gg \gg \frac{70}{110}$$

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(3) Find $P[R_1{G}_2]$ using multiplication rule:

$$P[R_1{G}_2]=P[R_1]\cdot P[G_2\mid R_1]\gg \gg \frac{4}{11}\cdot \frac{7}{10}\gg \gg \frac{28}{110}$$

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(4) Compare both sides:

Left side: $P[R_1{G}_2]=\frac{28}{110}$.

Right side:

$$P[R_1]\cdot P[G_2]=\frac{4}{11}\cdot \frac{70}{110}=\frac{28}{121}$$

But $\frac{28}{110}\ne \frac{28}{121}$ so $P[R_1{G}_2]\ne P[R_1]\cdot P[G_2]$ and they are not independent.

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Tree diagrams

Marble transferred, marble drawn

Marble transferred, marble drawn

Setup:

• Bin 1 holds five red and four green marbles.
• Bin 2 holds four red and five green marbles.

Experiment:

• You take a random marble from Bin 1 and put it in Bin 2 and shake Bin 2.
• Then you draw a random marble from Bin 2 and look at it.

Questions:

(a) What is the probability you draw a red marble?

(b) Supposing that you drew a red marble, what is the probability that a red marble was transferred?

Solution

Construct the tree diagram:

Identify sub-experiments, label events, compute probabilities:

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(a) Compute $P[D_R]$:

Add up leaf numbers for $D_R$ at leaf:

$$P[D_R]=\frac{25}{90}+\frac{16}{90}=\frac{41}{90}$$

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(b) Compute $P[T_R\mid D_R]$:

Conditional probability definition:

$$\begin{array}{c}P[T_R\mid D_R]=\frac{P[T_R{D}_R]}{P[D_R]}\\ \\ \gg \gg \frac{25/90}{41/90}\gg \gg \frac{25}{41}\end{array}$$

Interpretation: value of desired path over value of all possible paths.

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Counting

Counting teams with Cooper

Counting teams with Cooper

A team of 3 student volunteers is formed at random from a class of 40. What is the probability that Cooper is on the team?

Solution

There are $\left(\begin{array}{c}39\\ 2\end{array}\right)$ teams that include Cooper, and $\left(\begin{array}{c}40\\ 3\end{array}\right)$ teams in total. So we have:

$$P=\frac{39!}{2!37!}/\frac{40!}{3!37!}=\frac{3}{40}$$

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Haley and Hugo from 2 groups of 3

Haley and Hugo from 2 groups of 3

A UVA class has 40 students. Suppose the professor chooses 3 students on Wednesday at random, and again 3 on Friday. What is the probability that Haley is chosen today and Hugo on Friday?

Solution

(1) Count total outcomes:

• We have $\left(\genfrac{}{}{0ex}{}{40}{3}\right)$ possible groups chosen Wednesday.
• We have $\left(\genfrac{}{}{0ex}{}{40}{3}\right)$ possible groups chosen Friday.

Therefore $\left(\genfrac{}{}{0ex}{}{40}{3}\right)\times \left(\genfrac{}{}{0ex}{}{40}{3}\right)$ possible groups in total. (Product of possibilities.)

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(2) Count desired outcomes:

The possible groups of 3 that include Haley can be counted by counting the subgroups of 2 formed of the other students in Haley’s group.

• Therefore we have $\left(\genfrac{}{}{0ex}{}{39}{2}\right)$ groups that contain Haley.
• Similarly, we have $\left(\genfrac{}{}{0ex}{}{39}{2}\right)$ groups that contain Hugo.

Therefore we count $\left(\genfrac{}{}{0ex}{}{39}{2}\right)\times \left(\genfrac{}{}{0ex}{}{39}{2}\right)$ total desired outcomes.

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(3) Compute probability:

Let $E$ label the desired event. By the counting rule:

$$P[E]=\frac{|E|}{|S|}$$

Evaluate using our data:

$$\begin{array}{c}P[E]\gg \gg \frac{\left(\genfrac{}{}{0ex}{}{39}{2}\right)\times \left(\genfrac{}{}{0ex}{}{39}{2}\right)}{\left(\genfrac{}{}{0ex}{}{40}{3}\right)\times \left(\genfrac{}{}{0ex}{}{40}{3}\right)}\\ \\ \gg \gg {\left(\frac{\frac{39\cdot 38}{2!}}{\frac{40\cdot 39\cdot 38}{3!}}\right)}^2\gg \gg {\left(\frac{3}{40}\right)}^2\end{array}$$ Link to original

Counting VA license plates

Counting VA license plates

VA license plates have three letters (with no I, O, or Q) followed by four numerals. A random plate is seen on the road.

(a) What is the probability that the numerals occur in strictly increasing order?

(b) What is the probability that at least one number is repeated?

Solution

(a) Numerals in increasing order.

(1) Count total plates:

• Have $23\cdot 23\cdot 23$ options for letters.
• Have $10\cdot 10\cdot 10\cdot 10$ options for numbers.

Thus $23^3\cdot 10^4$ possible plates.

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(2) Count ways to have 4 numerals that occur in increasing order:

There are $\left(\genfrac{}{}{0ex}{}{10}{4}\right)$ ways to choose 4 distinct numerals from 10 options.

For each choice of four distinct numerals (i.e. for each combination), there is exactly one ordering that’s increasing.

Therefore, there are $\left(\genfrac{}{}{0ex}{}{10}{4}\right)$ ways to have 4 numerals that occur in increasing order.

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(3) Count ways to have a list of 3 letters (excluding I, O, Q).

There are 26 total letters, 3 are excluded, thus 23 options for each letter.

Repetition is allowed, thus we have $23\cdot 23\cdot 23=23^3$ total ways.

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(3) Compute probability:

Total count of desired plates (taking the product of possibilities):

$$23^3\cdot \left(\genfrac{}{}{0ex}{}{10}{4}\right)$$

Let $E$ label the event that a plate has increasing numerals. The counting formula for probability is:

$$P[E]=\frac{|E|}{|S|}$$

Evaluate using our data:

$$P[E]\gg \gg \frac{23^3\cdot \left(\genfrac{}{}{0ex}{}{10}{4}\right)}{23^3\cdot 10^4}\gg \gg \frac{\frac{10!}{4!6!}}{10000}\gg \gg \frac{21}{1000}$$

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(b) At least one number repeated.

“At least” is hard to work with! Lots of ways that can happen.

Try the complement event, which is much simpler: “no repetition”

Let $E^c$ be event that no numbers are repeated. All are distinct. Then $E$ is our desired event.

Count the possibilities:

$$|E^c|=23\cdot 23\cdot 23\cdot 10\cdot 9\cdot 8\cdot 7$$

The total number of plates is still $23^3\cdot 10^4$.

Therefore, license plates with at least one number repeated:

$$\begin{array}{c}|E|=|S|-|E|\\ \\ \gg \gg 23^3\cdot 10^4-23^3\cdot 10\cdot 9\cdot 8\cdot 7\gg \gg 60348320\end{array}$$

Desired outcomes over total outcomes:

$$\frac{|E|}{|S|}\gg \gg \frac{60348320}{23^3\cdot 10^4}\gg \gg 0.496$$ Link to original

Counting out 4 teams

Counting out 4 teams

A board game requires 4 teams of players. How many configurations of teams are there out of a total of 17 players if the number of players per team is 4, 4, 4, 5, respectively.

Solution

This is just the multinomial coefficient with this data:

$n$	$r_1$	$r_2$	$r_3$	$r_4$
17	4	4	4	5

So we have:

$$\#\text{teams}=\frac{n!}{r_1!r_2!r_3!r_4!}\gg \gg \frac{17!}{4!4!4!5!}\gg \gg 214414200$$

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