3.6 Summary of Probability Rules
We have learned the following probability rules so far:
 Complement Rule: [latex]P(\text{not }E) = 1  P(E)[/latex] or [latex]P(E) = 1  P(\text{not }E)[/latex].
 General Addition Rule: [latex]P(A \text{ or } B) = P(A) + P(B)  P(A \: \& \: B)[/latex].
 Special Addition Rule: [latex]P(A \text{ or } B) = P(A) + P(B)[/latex] if events A and B are mutually exclusive.
 Conditional Probability Rule: [latex]P(AB) = \frac{P(A \: \& \: B)}{P(B)}[/latex] for [latex]P(B) > 0[/latex].
 General Multiplication Rule: [latex]P(A \: \& \: B)=P(B) \times P(AB)[/latex] or [latex]P(A \: \& \: B)=P(A) \times P(BA)[/latex].
 Special Multiplication Rule: [latex]P(A \: \& \: B)=P(A) \times P(B)[/latex] if events A and B are independent.
Exercise: Application of Probability Rules
It is believed that there is an association between breast cancer and smoking. The following table summarizes results of an observational study of 200 females who are classified by their disease status and smoking status.
Smoker (S)  Nonsmoker (not S)  Total  

Breast Cancer (B)  10 (B & S)  30 (B & not S)  40 (B) 
Cancer Free (not B)  20 (not B & S)  140 (not B & not S)  160 (not B) 
Total  30 (S)  170 (not S)  200 
 What is the probability that a randomly selected female suffers from breast cancer?
 What is the probability that a randomly selected female is a smoker?
 What is the probability that a randomly selected female has breast cancer and she is a smoker?
 What is the probability that a randomly selected female has breast cancer given that she is a smoker?
 Are the events “Breast Cancer” and “Smoker” independent? Explain your answer by calculation.
 Interpret the conditional probability calculated in part (4) in several ways.
Show/Hide Answer
 What is the probability that a randomly selected female suffers from breast cancer?
It is very important to define the events in order to apply the probability rules. Let B = the event of suffering breast cancer, and S = the event of being a smoker. For this exercise, we want to compute [latex]P(B)[/latex]. Since we randomly pick one female and each female has the same chance of being selected, we can use the f/N rule.
[latex]P(B) = \frac{f}{N} = \frac{40}{200} = 0.2.[/latex]
 What is the probability that a randomly selected female is a smoker?
[latex]P(S) = \frac{f}{N} = \frac{30}{200} = 0.15.[/latex]
 What is the probability that a randomly selected female has breast cancer and she is a smoker?
[latex]P(B \: \& \: S) = \frac{f}{N} = \frac{10}{200} = 0.05.[/latex]
 What is the probability that a randomly selected female has breast cancer given that she is a smoker?
We want to compute P(BS). By the conditional probability rule
[latex]P(BS) = \frac{P(B \: \& \: S)}{P(S)} = \frac{10/200}{30/200} = \frac{1}{3} = 0.333.[/latex]
 Are the events “Breast Cancer” and “Smoker” independent? Explain your answer by calculation.
No, since [latex]P(BS) \neq P(B)[/latex]. The conditional probability of breast cancer given smoker is [latex]P(BS) = 0.333[/latex] and the unconditional probability of breast cancer [latex]P(B) = 0.2[/latex], which implies there is an association between smoking and higher rates of breast cancer.
We can also check whether [latex]P(B \: \& \: S)=P(B) \times P(S)[/latex]. Since [latex]P(B \: \& \: S)=\frac{10}{200} \neq (\frac{40}{200}) (\frac{30}{200}) = P(B) \times P(S)[/latex], the two events are NOT independent.
 Interpret the conditional probability calculated in part (4) in several ways.
We got [latex]P(BS)=0.333[/latex] in part (4), and this conditional probability can be interpreted in the following ways:

 If a randomly selected woman is a smoker, her probability of having breast cancer is 0.333.
 Given that a randomly selected woman smokes, she has a probability of 0.333 of having breast cancer.
 If you are a woman who smokes, then you have a probability of 0.333 of having breast cancer.
 33.3% of women who smoke have breast cancer.