CK-12 Probability and Statistics - Advanced

3.5. Additive and Multiplicative Rules http://www.ck12.org

And since the second toss is also a head,

P(B) =P(HH)+P(T H) 1 / 4 + 1 / 4 = 1 / 2

Now, what is the conditional probability? Here it is,

P(B|A) =

P(A∩B)

P(A)

=

1 / 4

1 / 2

=

1

2

What does this tell us? It tells us thatP(B) = 1 /2 andP(B|A) = 1 /2 also. Which means knowing that the first toss
resulted in a head does not affect the probability of the second toss. In other words,

P(B) =P(B|A)

When this occurs, we say that eventsAandBareindependent.

Condition of Independence

If eventBis independent of eventA, then the occurrence ofAdoes not affect the probability of the occurrence of
eventB. So we write,

P(B) =P(B|A)

Example:

The table below gives the number of physicists (in thousands) in the US cross classified by specialties(P 1 ,P 2 ,P 3 ,P 4 )
and base of practice(B 1 ,B 2 ,B 3 ). (Remark: The numbers are absolutely hypothetical and do not reflect the actual
numbers in the three bases.)

Suppose a physicist is selected at random. Is the event that the physicist selected is based in academia independent
of the event that the physicist selected is a nuclear physicist?

In other words, is the eventB1 independent ofP3?

TABLE3.3:

Industry

(B 1 )

Academia

(B 2 )

Government

(B 3 )

Total

General Physics

(P 1 )

10. 3 72. 3 11. 2 93. 8

Semiconductors

(P 2 )

11. 4 0. 82 5. 2 17. 42