# Conditional Probability: Bayesian Cause/Effect Question

The probability that a randomly chosen male has a circulation problem is 0.25. Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem. What is the conditional probability that a male has a circulation problem, given that he is a smoker?

Let:

C = Circulation Problem

S = smoker.

The problem is asking you to solve for $\Pr(C \mid S)$. I thought I could solve problem by using a table and I completed it below. I interpreted the statements as such:

1: "a randomly chosen male has a circulation problem is 0.25": $\Pr(C)=0.25$

2: "Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem.": $\Pr(C \cap S) = 2x$ and $\Pr(C' \cap S) = x$

\begin{array} {|c|c|c|c} \hline & C & C' & \\ \hline S& 2x& x&\\ \hline S'& & &\\ \hline & 0.25 & &1\\ \hline . \end{array}

Which becomes:

\begin{array} {|c|c|c|c} \hline & C & C' & \\ \hline S& 2x& x & 3x\\ \hline S'& & &\\ \hline & 0.25 & 0.75 &1\\ \hline . \end{array}

So I did $\Pr(C\mid S) = \cfrac{\Pr(C \cap S)}{\Pr(S)}=\cfrac{2x}{3x}=2/3$ which is wrong. The answer is 2/5!

Am I incorrectly interpreting statement 2 above as an intersection when instead it is a conditional probability? In other words, should I be writing $\Pr(S\mid C)=2x$ and $\Pr(S\mid C')=x$?

Thank you in advance.

• Your point $2$ should be $P(S|C)$ and $P(S|C')$, not $P(S\cap C)$ and $P(S\cap C')$. "A male with circulation problem has some probability to ..." means "$\ldots |C$", rather than "$\ldots \cap C$". – Arthur May 20 '13 at 22:46
• So the relation you get from the sentence "Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem." is $\frac{P(S\cap C)}{0.25} = 2\frac{P(S\cap C')}{0.75}$. – Arthur May 20 '13 at 22:51
• Thank you @Arthur. How do you know interpret it as conditional probability instead of intersection as I did? Is it because it is stated as "twice as likely" and not as fact? – user1527227 May 20 '13 at 22:54
• You have to read that from the text. If what they're saying is "Take a man who has property $A$", what they're really saying is "Assume $A$ has already ocurred", which is what $|A$ means. Another way to look at it (if you're up for the abstract challenge) is by symmetry. Conditional probabilities are asymmetrical in nature, whereas intersections are symmetrical. Look at the distinction between "The probability that a man with circulation problems is also smoking is high" and the sentence "The probability that a man both smokes and has circulation problems is high". – Arthur May 20 '13 at 23:01

## 1 Answer

I was able to answer/understand this question with Arthur's comments. In this question, I had the problem of distinguishing between a conditional and nonconditional probability question. This is also a good post to view for anyone who has problems:

How do we identify a probability problem as a conditional probability problem?

The proper way to mathematically represent the statements are as follows:

1: $\Pr(C) = 0.25$: The probability that a randomly chosen male has a circulation problem is 0.25.

2: $\Pr(S \mid C) = 2 * \Pr(S \mid C')$: Males who have a circulation problem are twice as likely to be smokers as those who do not have a circulation problem. From this statement we can also infer:

3: $\Pr(C \mid S):$What is the conditional probability that a male has a circulation problem, given that he is a smoker?

In this question, we are asked for a specific group within a group which is hint to conditional probability. This problem could be diagrammed as a tree where the first node would be whether the person has a circulation problem (C or C'). Then the C and C' nodes would each have two branches come from them: S and S'.

In other words, whether you're a smoker or have a circulation problem are not equal status. One is higher than another if viewed as a hierarchy.

From definition of conditional probability:

$\Pr(C \mid S) = \cfrac{\Pr(C \cap S)}{\Pr(S)} = \cfrac{\Pr(C \cap S)}{Pr(S \cap C) + \Pr(S \cap C')}=\cfrac{0.25*2x}{0.25*2x + 0.75*x}=2/5$

Notice that in conditional probability problems, they can be both represented as a tree diagram or a venn box diagram.