Probability: Conditional In-Depth (Dependent Events)

Introduction

In the Probability Rules and Terms page, we introduced the concept of dependent events. The definition is below:

Events are dependent if the occurrence of one event affects the probability of the other occurring

One way to help identify whether or not the problem deals with dependent events is to look for the key word "given." The probability of A given B is dependent because the occurrence of B affects the probability of A occurring. A real-life example of this would be: "What is the probability that team A wins given that they are leading by 20 points?"

Notation

The probability of event A occurring given that event B has already occurred is written with the notation P(A | B) and read as "the probability of A given B."

Formula

$$P(A\mid B)=\frac{P(A \: \cap\: B)}{P(B)}$$    OR   $$P(A\mid B)=\frac{P(A \: and\: B)}{P(B)}$$

Example

The following is data collected from a survey of 200 random people.

 Favorite Pet / Favorite Sport Basketball Football Baseball Sum Dog 47 54 10 111 Cat 25 22 31 78 Rabbit 2 3 6 11 Sum 74 79 47 200

A random person's survey results were viewed. What is the probability that this individual chose a dog as their favorite pet given that they chose baseball as their favorite sport?

 Solution A (Using Mathematics) Solution B (Using Logic) P(Dog | Baseball) = P(Dog and Baseball) / P(Baseball) P(Dog | Baseball) = (10/200) / (47/200)  = (10/47) = .213 Through inspection, we know that there are own only 47 individuals who like baseball. If we took those 47 into a separate room, then randomly picked a person, what would be the probability that this person liked dogs the most? Well of the 47 individuals, 10 like dogs. Thus, the probability a person likes a dog the most, given that they like baseball the most, is simply 10/47 or 0.213.

### Multiplication Rule for Related Events

If two events, A and B, are NOT independent events, then the probability of A and B is equal to the probability of A times the probability of B given A.

 $$\cap$$ Noation "and" Notation $$P(A\: \cap\: B)=P(A) P(B\mid A)$$ $$P(A\: and\: B)=P(A) P(B\mid A)$$ Similarly, the following formulas also apply: $$P(A\: \cap\: B)=P(B) P(A\mid B)$$ $$P(A\: and\: B)=P(B) P(A\mid B)$$

Example:
A person is presented with a bag full of 10 marbles. 5 are red and 5 are green. The person is asked to take one marble out, record the marble's color and then set it aside. The person is then asked to draw another marble and record its color. Calculate the probability that both marbles are green.
Solution:
P(both green) = P(1st green and 2nd green) = P(1st green)P(2nd green | 1st green)
P(1st green) = $$\frac{1}{2}$$
P(2nd green | 1st green) = $$\frac{4}{9}$$     Note: After the person removes a green marble, there are 9 marbles remaining. Of the 9, only 4 are green.
P(both green) = P(1st green)P(2nd green | 1st green) = (1/2)(4/9) = 2/9

Tree Diagrams

Tree Diagrams are a powerful tool for conditional probability. To see examples of problems being solved with this tool, click on the Tree Diagram help page.