Conditional Probability

Terms

Conditional probability – Probability of one event given another has occurred.

The conditional probability of A given B is denoted by \(P(A | B)\)

How do we find the conditional probabilty, \(P(A | B)\)

\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]

In general, \(P(A | B) \neq P(B | A)\).

Example

Consider the probability of a flight departing on time \(P(D)\) is 0.83, and the probability of a flight arriving on time \(P(A)\) is 0.82. The probability that a flight departs and arrives on time \(P(D \cap A)\) is 0.78.

Find the probability

• the flight arrives on time given it departed on time
• the flight departed on time given it arrived on time

The first question asks for \(P(A | D)\), which we can find using \(\frac{P(A \cap D)}{P(D)} = \frac{0.78}{0.83} = 0.94\).

The second question asks us it in reverse: \(P(D | A) = \frac{0.78}{0.82} = 0.95\).

Notice the two are not necessarily the same.

Example

There are 20 balls in a box, either red or green, either rubber or plastic. If we know that 12 balls are red and 9 balls are rubber and 4 balls are green and rubber:

• Find the number of red plastic balls
• If one ball is red, what is the probability that this ball is plastic?
• If one ball is green, what is the probability that this ball is plastic?

First question:

# of plastic balls = 20 - 9 = 11

# of plastic and green balls = (20 - 12) - 4 = 4

# of plastic and red balls = 11 - 4 = 7

Second question:

\(P(P | R) = \frac{P(P \cap R)}{P(R)}\)

\(\frac{7}{12}\)

Third question:

\(P(P | G) = \frac{P(P \cap G)}{P(G)}\)

\(\frac{4}{20 - 12}\)

\(\frac{1}{2}\)

Independent Events

When two events A and B are independent:

\(P(A \cap B) = P(A) * P(B)\)

Also complements like A’ and B’ are independent of each other:

• A’ and B are independent
• A and B’ are independent
• A’ and B’ are independent

Example

Insurance companies assume that there is a difference between gender and your likelihood of getting into an accident which is why women generally have lower insurance rates than men. We did a study to see the number of accidents that have occurred according to gender. We found:

• 60% of the population was male.
• 86% of the population was either male OR got into an accident.
• 35% of the population are accident free.

Does this study indicate that the likelihood of one to get into an accident depends on gender?

Suppose we have two events, A and B:

• A is the event a person gets into an accident
• B is the event a person is male

The question is, are these two events indepdendent?

\(P(A) = 1 - 0.35 = 0.65\) \(P(B) = 0.6\) \(P(A \cap B) = 0.39\). This can be determined by \(P(A) + P(B) - P(A \cup B)\).

Notice that \(P(A \cap B) = P(A) * P(B)\). This means that events A and B are independent. In other words, the event a person gets into an accident does not depend on the gender of the person.

Product Rule

\(P(A \cap B) = P(A) * P(B|A)\)

• This is just \(P(A) * P(B)\) when A and B are independent