CK-12 Probability and Statistics - Advanced

http://www.ck12.org Chapter 3. An Introduction to Probability

TABLE3.3:(continued)

Industry

(B 1 )

Academia

(B 2 )

Government

(B 3 )

Total

Nuclear Physics

(P3)

1. 25 0. 32 34. 3 35. 87

Astrophysics

(P 4 )

0. 42 31. 1 35. 2 66. 72

Total 23. 37 104. 54 85. 9 213. 81

Figure:A table showing the number of physicists in each specialty (thousands). This data is hypothetical.

Solution:

The problem may appear a little difficult at first but it is actually much simpler, especially, if we make use of the
condition of independence. All we need to do is to calculateP(B 1 |P 3 )andP(B 1 ). If those two probabilities are
equal, then the two eventsB1 andP3 are indeed independent. Otherwise, they are dependent. From the table we
find,

P(B 1 ) =

23. 37

213. 81

= 0. 109

And

P(B 1 |P 3 ) =

P(P 3 ∩B 1 )

P(P 3 )

=

1. 25

35. 87

= 0. 035

Thus,P(B 1 |P 3 ) 6 =P(B 1 )and so the eventB1 is dependent on the eventP3. This lack of independence results from
the fact that the percentage of nuclear physicists who are working in the industry( 3 .5%)is not the same as the
percentage of all physicists who are in the industry( 10 .9%).

Lesson Summary

1. TheAdditive Rule of Probabilitystates that the union of two events can be found by adding the probabilities
   of each event and subtracting the intersection of the two events, orP(A∪B) =P(A)+P(B)−P(A∩B).


2. IfA∩Bcontains no simple events, thenAandBaremutually exclusive. Mathematically, this meansP(A∪
   B) =P(A)+P(B).


3. TheMultiplicative Rule of ProbabilitystatesP(A∩B) =P(B)P(A|B).


4. If eventBisindependentof eventA, then the occurrence ofAdoes not affect the probability of the occurrence
   of eventB. Mathematically,P(B) =P(B|A).