Conditional Probability

Sometimes knowledge that one event has occurred will change the probability that another event will occur. Let \(A\) and \(B\) be two events with \(P(B)>0\). Then \[P(A|B)=\frac{P(A \cap B)}{P(B)}\] where \(P(A|B)\) is read as “the probability of \(A\) given \(B\)” and tells us the probability of \(A\) occurring given that \(B\) has occurred.

Suppose I randomly choose a card from a standard deck. The probability that the card I chose is a Queen is \(\frac{4}{52}\) since there are 52 cards and 4 of them are Queens.

Now, suppose I randomly choose a card from a standard deck and I tell you it is a face card. Now the probability that the card chosen is a Queen is \(\frac{4}{12}\) since there are 12 face cards and 4 of them are Queens. Using the formula for conditional probability

\[P(\text{Queen }|\text{ face})=\frac{P(\text{Queen } \cap \text{ face})}{P(\text{face})}=\frac{\frac{4}{52}}{\frac{12}{52}}=\frac{4}{12}\]

In the first case, the sample space consisted of all 52 cards. In the second case, the sample space was reduced to the 12 face cards. So you can think of conditional probability as reducing the sample space.

Example 1:

Sixty people were asked for their gender and political party.

Democrat Republican Independent Total
Male 10 5 5 20
Female 15 15 10 40
Total 25 20 15 60

Find the probability that a randomly selected person is female given that the person is a Democrat.

\(P(\text{female|Democrat}) = \frac{P(\text{female } \cap \text{ Democrat})}{P(\text{Democrat})} = \frac{\frac{15}{60}}{\frac{25}{60}}=\frac{15}{25}\)

Multiplication Rule

\[ P(A \cap B) = P(A|B)P(B)\]

Example 2:

Seat belts are worn by drivers in 65% of reported accidents. When seat belts were worn, 8% of the accidents were fatal. When seat belts were not worn, 20% of the accidents were fatal.

  • What is the probability that in an accident the driver is wearing a seat belt and is killed?
  • What is the probability that in an accident the driver is not wearing a seat belt and is killed?
  • What is the probability that in an accident the driver is not wearing a seat belt and is not killed?
  • What is the probability that in an accident the driver is killed?

We can organize this information in a tree diagram.
Define the following events: \(S\) = wearing seatbelt and \(F\) = fatal accident.


The first three answers can be found directly from the tree diagram

  • \(P(S \cap F) = 0.052\)
  • \(P(S^c \cap F) = 0.07\)
  • \(P(S^c \cap F^c) = 0.28\)

The last answer is a little different from the problems above. For this one we need to add two probabilities from our tree. Notice that there are two ways to be in a fatal car accident: either you are wearing your seatbelt and you are in a fatal car accident (\(S \cap F\)) or you are not wearing your seatbelt and you are in a fatal car accident (\(S^c \cap F\)). These two events are mutually exclusive, so we can add their probabilities:

  • \(P(F) = 0.052 + 0.07 = 0.122\)

Independent Events

Two events \(A\) and \(B\) are independent if \(P(A|B)=P(A)\), that is if knowledge that one event has occurred doesn’t affect the probability of the other event occurring. Otherwise, \(A\) and \(B\) are dependent events.

Example 3:

Sixty people were asked for their gender and political party.

Democrat Republican Independent Total
Male 10 5 5 20
Female 15 15 10 40
Total 25 20 15 60
  • Are the events “female” and “Democrat” independent?

  • No since \(P(\text{female|Democrat}) \ne P(\text{female})\)

Multiplication Rule (Independent)

if \(A\) and \(B\) are independent events then the multiplication rule simplifies to \(P(A \cap B) = P(A)P(B)\)

Example 4:

A company that explores for oil is considering two new sites, site A and site B. The probability of finding oil at site A is 0.6 and at site B is 0.4. The events “find oil at site A” and “find oil at site B” are independent. Find the probability that

  • oil is found at both sites
  • oil is found at site A but not at site B
  • oil is found at only one site

  • \(P(A \cap B) = P(A)P(B) = 0.6 \cdot 0.4 = 0.24\)
  • \(P(A \cap B^c) = P(A)P(B^c) = 0.6 \cdot (1-0.4) = 0.36\)
  • \(P(A \cap B^c) + P(A^c \cap B)= P(A)P(B^c) + P(A^c)P(B) = 0.36 + 0.16 = 0.52\)

Bayes’ Rule

Suppose \(B_1\) and \(B_2\) are mutually exclusive events with \(P(B_1)+P(B_2)=1\). Then, for any event \(E\),

\[P(B_1|E) = \frac{P(B_1 \cap E)}{P(E)} = \frac{P(E|B_1)P(B_1)}{P(E|B_1)P(B_1)+P(E|B_2)P(B_2)}\]

Essentially, Bayes’ Rule allows us to calculate \(P(B_1|E)\) from \(P(E|B_1)\).

Example 5:

Suppose the incidence of Lyme disease in a population is 207 per 100,000 people. There is a test for Lyme disease that is accurate 93.7% of the time for people that actually have Lyme disease and is accurate 97% of the time for people that do not have Lyme disease.

If a randomly selected person tests positive for Lyme disease, what is the probability that the person actually has Lyme disease?

Let \(L\) denote the event that a person has Lyme disease. Let \(+\) denote the event of a positive test. Let \(-\) denote the event of a negative test.

We are given that

  • \(P(L) = \frac{207}{100000} = 0.00207\)
  • \(P(+|L) = 0.937\)
  • \(P(-|L^c) = 0.97\)

Putting this into a tree diagram gives


Now applying Bayes’ rule and pulling the appropriate values off the tree diagram

\[P(L|+)=\frac{P(L \cap +)}{P(+)}=\frac{0.00194}{0.00194+0.02994}=0.06\]

Note that there are two ways to get a positive test so to find \(P(+)\) we had to add values from two branches of the tree.