Chapter 2 Event, Sample Space and Probability (Section on Feb 3rd)

Basic Definition

Definition 2.1 (Sample Space) The sample space (also known as the outcome space) of an experiment is simply the set of all possible outcomes associated with it. We often denote it be $\mathcal{S}$.

Definition 2.2 (Event) Any subset $E$ of sample space $\mathcal{S}$ is known as an event.

Definition 2.3 (Probability) A probability is a number between 0 and 1 that we attach to each element of the sample space. Informally, that number simply describes he chance of that event happening. A probability of 1 means that the event will happen for sure. A probability of 0 means that we are talking about an impossible event. The number in between represent various degrees of certainy about he occurence of the event.

Classical Probability Model

Assume all outcomes of a certain experiment are equally likely to happen. Then the probability of a certain event is just the number of outcomes satisfies the criteria of the event, devided by the total number of outcomes. This model gives us a way to calculate probability.

• Step 1: Count the number of total outcomes, denote by $N$.

• Step 2: Count the number of outcomes satisfies the criteria of a certain event, denote by $n$.

• Step 3: The probability of the event is then $\frac{n}{N}$.

Exercises

Exercise 2.1 If a couple plans to have six children, what is the probability that they will have at least one girl?

Proof. Solve by steps.

Step 1: Number of total outcomes: $2\times 2\times 2\times 2\times 2\times 2=2^6=64$.

Step 2: Number of outcomes contains at least one girl. This number is hard to compute, because we need to consider the case that there are $1, 2,\cdots,6$ girls. Therefore, we consider the inverse. The inverse of at lease one girl is all children are boys. Only one out of the 64 outcomes satisfies this. So the number of outcomes satisfies at least one girl is: $$\begin{equation} 64-1=63 \tag{2.1} \end{equation}$$

Step 3: Divide: $\frac{63}{64}\approx0.9844$. So the probability is 0.9844.

If the outcomes of a certain event is hard to get, consider the inverse of it.

Exercise 2.2 Suppose that 5 fair coins are tossed.

1. Describe the sample space for this experiment.

2. What is the probability that all 5 tosses will have the same outcome?

3. What is the probability that at least 3 heads appear in 5 tosses?

Proof. (a) List all possible outcomes.

• All head: $\{H,H,H,H,H\}$;

• Four heads: $\{T,H,H,H,H\}$, $\{H,T,H,H,H\}$, $\{H,H,T,H,H\}$, $\{H,H,H,T,H\}$, $\{H,H,H,H,T\}$;

• Three heads: $\{T,T,H,H,H\}$, $\{T,H,T,H,H\}$, $\{T,H,H,T,H\}$, $\{T,H,H,H,T\}$, $\{H,T,T,H,H\}$, $\{H,T,H,T,H\}$, $\{H,T,H,H,T\}$, $\{H,H,T,T,H\}$, $\{H,H,T,H,T\}$, $\{H,H,H,T,T\}$;

• Two heads: $\{T,T,T,H,H\}$, $\{T,T,H,T,H\}$, $\{T,T,H,H,T\}$, $\{T,H,T,T,H\}$, $\{T,H,T,H,T\}$, $\{T,H,H,T,T\}$, $\{H,T,T,T,H\}$, $\{H,T,T,H,T\}$, $\{H,T,H,T,T\}$, $\{H,H,T,T,T\}$;

• One heads: $\{T,T,T,T,H\}$, $\{T,T,T,H,T\}$, $\{T,T,H,T,T\}$, $\{T,H,T,T,T\}$, $\{H,T,T,T,T\}$;

• No head: $\{T,T,T,T,T\}$.

Total number of outcomes: $$\begin{equation} 1+5+10+10+5+1=32 \tag{2.2} \end{equation}$$

2. Number of outcomes satisfies criteria is 2, so the probability is $\frac{2}{32}=0.0625$.

3. Number of outcomes satisfies criteria is $10+5+1=16$, so the probability is $\frac{16}{32}=0.5$.

When count number of total outcomes, self-define some rules to help preventing over or under counting.

Exercise 2.3 A total of 28 percent of American males smoke cigarettes, 7 percent smoke cigars, and 5 percent smoke both cigars and cigarettes.

1. What percentage of male smokes neither cigars nor cigarettes?

2. What percentage smoke cigars but not cigarettes?

Proof. Draw a Venn Diagram to solve this problem.

Define Event A: American males smoke cigarettes; Event B: American males smoke cigars; and Set $\mathcal{S}$: all American males.

Suppose there are $N$ American males, then $Pr(A)=28\%=0.28$, $Pr(B)=7\%=0.07$ and $Pr(A\cap B)=5\%=0.05$.

1. It asks $Pr((A\cup B)^c)$.

$$\begin{equation} Pr((A\cup B)^c)=1-Pr(A\cup B)=1-Pr(A)-Pr(B)+Pr(A\cap B)=1-0.028-0.07+0.05=0.7=70\% \tag{2.3} \end{equation}$$

2. It asks $Pr(B\setminus (A\cap B))$. $$\begin{equation} Pr(B\setminus (A\cap B))=Pr(B)-Pr(A\cap B)=0.07-0.05=0.02=2\% \tag{2.4} \end{equation}$$

Exercise 2.4 Three cards are drawn from a deck of cards. What is the probability that the first two cards are J and the third card is 10?

Proof. - Step 1: Choose first card, which is a J: $\frac{4}{52}$

• Step 2: Choose second card, which is also a J: $\frac{3}{51}$

• Step 3: Choose third card, which is a 10: $\frac{4}{50}$

The probability is then: $$\begin{equation} \frac{4}{52}\times\frac{3}{51}\times\frac{4}{50}\approx 3.6199\times 10^{-4} \end{equation}$$

Reference

• Rule of Product, sometimes also called Multiplication Principle.

• Permutation

• Combination