Examples

Lucia is Host or Player

The professor chooses three students at random for a game in a class of 40, one to be Host, one to be Player, one to be Judge. What is the probability that Lucia is either Host or Player?

Solution

(1) Set up the probability model:

Label the students $1$ to $40$. Write $L$ for Lucia’s number.

Outcomes: assignments such as $(H,P,J)=(\mathrm{2,5,8})$. These are ordered triples with distinct entries in $1,2,\dots ,40$.

Sample space: $S$ is the collection of all such distinct triples

Events: any subset of $S$

Probability measure: assume all outcomes are equally likely, so $P[(i,j,k)]=P[(r,l,p)]$ for all $i,j,k,r,l,p$

In total there are $40\cdot 39\cdot 38$ triples of distinct numbers.

Therefore $P[(i,j,k)]=\frac{1}{40\cdot 39\cdot 38}$ for any specific outcome $(i,j,k)$.

Therefore $P[A]=\frac{|A|}{40\cdot 39\cdot 38}$ for any event $A$. (Recall $|A|$ is the number of outcomes in $A$.)

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(2) Define the desired event:

We want to find $P[\text{“Lucia is Host or Player”}]$. Define $A=\text{“Lucia is Host”}$ and $B=\text{“Lucia is Player”}$. Thus:

$$A=\{(L,j,k)|\text{any }j,k\},B=\{(i,L,k)|\text{any }i,k\}$$

So, in this notation, we seek $P[A\cup B]$.

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(3) Compute the desired probability:

Importantly, $A\cap B=\varnothing$ (mutually exclusive). There are no outcomes in $S$ in which Lucia is both Host and Player.

By additivity, we infer $P[A\cup B]=P[A]+P[B]$.

Now compute $P[A]$. There are $39\cdot 38$ ways to choose $j$ and $k$ from the students besides Lucia. Therefore $|A|=39\cdot 38$. Therefore:

$$P[A]\gg \gg \frac{|A|}{40\cdot 39\cdot 38}\gg \gg \frac{39\cdot 38}{40\cdot 39\cdot 38}\gg \gg \frac{1}{40}$$

Now compute $P[B]$. It is similar: $P[B]=\frac{1}{40}$.

Finally compute that $P[A]+P[B]=\frac{1}{20}$, so the answer is:

$$P[A\cup B]\gg \gg P[A]+P[B]\gg \gg \frac{1}{20}$$

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”.

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(2) Define the desired event:

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

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

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(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:

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