W01 - Examples

Events and outcomes

Coin flipping

Coin flipping

Flip a fair coin two times and record both results.

• Outcomes: sequences, like $HH$ or $TH$.

• Sample space: all possible sequences, i.e. the set $S=\{HH,HT,TH,TT\}$.

• Events: for example:

  • $A=\{HH,HT\}=\text{“first was heads”}$
  • $B=\{HT,TH\}=\text{“exactly one heads”}$
  • $C=\{HT,TH,HH\}=\text{“at least one heads”}$

With this setup, we may combine events in various ways to generate other events:

Complex events: for example: $A\cap B=\{HT\}$, or in words:

$$\text{“first was heads”}\text{AND}\text{“exactly one heads”}=\text{“heads-then-tails”}$$

Notice that the last one is a complete description, namely the outcome $HT$.
$A\cup B=\{HH,HT,TH\}$, or in words:

$$\begin{array}{c}\text{“first was heads”}\text{OR}\text{“exactly one heads”}\\ =\text{“starts with heads, else it’s tails-then-heads”}\end{array}$$ Link to original

Coin flipping: counting subsets

Coin flipping: counting subsets

Flip a fair coin five times and record the results.

How many elements are in the sample space? (How big is $S$?)
How many events are there? (How big is $\mathcal{F}$?)

Solution

There are $2^5=32$ possible sequences, so $|S|=32$.

To count the number of possible subsets, consider that we have 32 distinct items, and a subset is uniquely determined by the binary information – for each item – of whether it is in or out. Thus there are $2^{32}$ possibilities. So $|\mathcal{F}|=2^{32}$.

Link to original

iPhones and iPads

iPhones and iPads

At Mr. Jefferson’s University, 25% of students have an iPhone, 30% have an iPad, and 60% have neither.

What is the probability that a randomly chosen student has some iProduct? (Q1)

What about both? (Q2)

Solution

(1) Set up the probability model:

A student is chosen at random:

Outcomes are chosen students.

The sample space $S$ is the set of all students. Events are subsets of $S$.

Write $O=\text{“has iPhone”}$ and $A=\text{“has iPad”}$ (regarding the chosen student).

All students are equally likely to be chosen. Therefore $P[E]=\frac{|E|}{|S|}$ for any event $E$. Therefore $P[O]=0.25$ and $P[A]=0.30$.

Furthermore, $P[O^c{A}^c]=0.60$. This states that 60% have “not iPhone AND not iPad”.

---

(2) Define the desired event:

Q1: $\text{desired event}=O\cup A$

Q2: $\text{desired event}=OA$

---

(3) Compute the probabilities:

We do not know that $O$ and $A$ are exclusive.

We could try inclusion-exclusion:

$$P[O\cup A]=P[O]+P[A]-P[OA]$$

We know $P[O]=0.25$ and $P[A]=0.30$. So this formula, with given data, RELATES Q1 and Q2. It does not solve either one by itself.

We have not yet used the information that $P[O^c{A}^c]=0.60$.

To use this, simplify it with De Morgan’s Laws:

$$P[O^c{A}^c]\gg \gg P[(O\cup A{)}^c]\gg \gg 1-P[O\cup A]$$

Therefore:

$$P[O^c{A}^c]=0.60\gg \gg P[O\cup A]=0.40$$

We have answered Q1. Recall that inclusion-exclusion relates Q1 and Q2 and solve to answer Q2:

$$\begin{array}{c}P[O\cup A]=P[O]+P[A]-P[OA]\\ \\ \gg \gg 0.40=0.25+0.30-P[OA]\\ \\ \gg \gg P[OA]=0.15\end{array}$$Link to original

Conditional probability

Coin flipping: at least 2 heads

Coin flipping: at least 2 heads

Flip a fair coin 4 times and record the outcomes as sequences, like HHTH.

Let $A_{\ge 2}$ be the event that there are at least two heads, and $A_{\ge 1}$ the event that there is at least one heads.

First let’s calculate $P[A_{\ge 2}]$.

Define $A_2$, the event that there were exactly 2 heads, and $A_3$, the event of exactly 3, and $A_4$ the event of exactly 4. These events are exclusive, so:

$$P[A_{\ge 2}]=P[A_2\cup A_3\cup A_4]\gg \gg P[A_2]+P[A_3]+P[A_4]$$

Each term on the right can be calculated by counting:

$$\begin{array}{rl}P[A_2]& =\frac{|A_2|}{2^4}\gg \gg \frac{\left(\genfrac{}{}{0ex}{}{4}{2}\right)}{16}\gg \gg \frac{6}{16}\\ & \\ P[A_3]& =\frac{|A_3|}{2^4}\gg \gg \frac{\left(\genfrac{}{}{0ex}{}{4}{3}\right)}{16}\gg \gg \frac{4}{16}\\ & \\ P[A_4]& =\frac{|A_4|}{2^4}\gg \gg \frac{\left(\genfrac{}{}{0ex}{}{4}{4}\right)}{16}\gg \gg \frac{1}{16}\end{array}$$

Therefore, $P[A_{\ge 2}]=\frac{11}{16}$.

Now suppose we find out that “at least one heads definitely came up”. (Meaning that we know $A_{\ge 1}$.) For example, our friend is running the experiment and tells us this fact about the outcome.

Now what is our estimate of likelihood of $A_{\ge 2}$?

The formula for conditioning gives:

$$P[A_{\ge 2}\mid A_{\ge 1}]=\frac{P[A_{\ge 2}\cap A_{\ge 1}]}{P[A_{\ge 1}]}$$

Now $A_{\ge 2}\cap A_{\ge 1}=A_{\ge 2}$. (Any outcome with at least two heads automatically has at least one heads.) We already found that $P[A_{\ge 2}]=\frac{11}{16}$. To compute $P[A_{\ge 1}]$ we simply add the probability $P[A_1]$, which is $\frac{4}{16}$, to get $P[A_{\ge 1}]=\frac{15}{16}$.

Therefore:

$$P[A_{\ge 2}\mid A_{\ge 1}]=\frac{11/16}{15/16}\gg \gg \frac{11}{15}$$ Link to original

Multiplication: flip a coin, then roll dice

Multiplication: flip a coin, then roll dice

Flip a coin. If the outcome is heads, roll two dice and add the numbers. If the outcome is tails, roll a single die and take that number. What is the probability of getting a tails AND a number at least 3?

Solution

(1) This “two-stage” experiment lends itself to a solution using the multiplication rule for conditional probability.

Label the events of interest.

Let $H$ and $T$ be the events that the coin showed heads and tails, respectively.

Let $A_1,\dots ,A_{12}$ be the events that the final number is $1,\dots ,12$, respectively.

The value we seek is $P[T{A}_{\ge 3}]$.

---

(2) Observe known (conditional) probabilities.

We know that $P[H]=\frac{1}{2}$ and $P[T]=\frac{1}{2}$.

We know that $P[A_5\mid T]=\frac{1}{6}$, for example, or that $P[A_2\mid H]=\frac{1}{36}$.

---

(3) Apply multiplication rule:

$$P[T{A}_{\ge 3}]=P[T]\cdot P[A_{\ge 3}\mid T]$$

We know $P[T]=\frac{1}{2}$ and can see by counting that $P[A_{\ge 3}\mid T]=\frac{2}{3}$.

Therefore $P[T{A}_{\ge 3}]=\frac{1}{3}$.

Link to original

Multiplication: draw two cards

Multiplication: draw two cards

Two cards are drawn from a standard deck (without replacement).

What is the probability that the first is a 3, and the second is a 4?

Solution

This “two-stage” experiment lends itself to a solution using the multiplication rule for conditional probability.

(1) Label events:

• Write $T$ for the event that the first card is a 3.
• Write $F$ for the event that the second card is a 4.

We seek $P[TF]$. We will use the multiplication rule:

$$P[TF]=P[T]\cdot P[F\mid T]$$

---

(2) Compute probabilities:

We know $P[T]=\frac{4}{52}$. (Does not depend on the second draw.)

For the conditional probability, note that if the first is a 3, then there are four remaining 4s and 51 remaining cards. Therefore:

$$P[F\mid T]=\frac{4}{51}$$

---

(3) Apply multiplication rule:

$$\begin{array}{c}P[TF]=P[T]\cdot P[F\mid T]\\ \\ \\ P[TF]=\frac{4}{52}\cdot \frac{4}{51}\gg \gg \frac{4}{13\cdot 51}\end{array}$$ Link to original

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.

What is the probability that the marble you look at is red?

Solution

(1) Label events:

• Event $T_R$: a red marble is transferred.
• Event $T_G$: a green marble is transferred.
• Event $D_R$: a red marble is drawn from Bin 2.
• Event $D_G$: a green marble is drawn from Bin 2.

Answer will be $P[D_R]$.

---

(2) Apply Division into Cases:

General formula: $P[B]=P[A]\cdot P[B\mid A]+P[A^c]\cdot P[B\mid A^c]$

Insert our labels, $A\to T_R$ and $B\to D_R$ and $A^c\to T_G$. Obtain:

$$P[D_R]=P[T_R]\cdot P[D_R\mid T_R]+P[T_G]\cdot P[D_R\mid T_G]$$

---

(3) Plug in data and compute:

• Know $P[T_R]=5/9$.
• Know $P[T_G]=4/9$.
• Know $P[D_R\mid T_R]=5/10$.
• Know $P[D_R\mid T_G]=4/10$.

Therefore:

$$P[D_R]\gg \gg \frac{5}{9}\cdot \frac{5}{10}+\frac{4}{9}\cdot \frac{4}{10}\gg \gg \frac{41}{90}$$

Link to original