Operations with Probabilities

Just as events can be treated as sets, so can probabilities of an event (in a sense). The formulas used to calculate the probability of unions, intersections and complements of events are similar to the ones used for sets. Given that event A and event B are subsets of the sample space S, the following rules apply:

(1) Union of events A and B

P (A È B) = P(A) + P(B) - P(A Ç B)

where P (A È B) is the probability that either events A or B occur.
P (A Ç B) is the probability that both events A and B occur.

If events A and B are mutually exclusive,
then P(A Ç B) = P(Æ) = 0, and

P (A È B) = P(A) + P(B)

In general, if a set of events A₁ , A₂ , A₃,....., A_n are mutually exclusive,

then

P (A1 È A₂ È ......... È An ) = P(A₁ ) + P(A₂ ) + ...... + P(A_n )

(2) Intersection of events A and B

From the definition of conditional probability,

P(A|B) = P(A Ç B) / P(B) and P(B|A) = P(A Ç B) / P(A).

Therefore,

P(A Ç B) = P(A) P(B|A) = P(B) P(A|B)

If A and B are independent events, then

P(A Ç B) = P(A) P(B)

In general, if a set of events A₁ , A₂ , ....., A_n are independent, then

P (A₁ Ç A₂ Ç ......... Ç A_n ) = P(A₁ ) P(A₂ ) ...... P(A_n )

(3) Complement of event A

Since A È A' = S, and the event A and its complement are mutually exclusive,

P(S) = P(A È A') = P(A) + P(A') = 1
P(A') = 1 - P(A)

EX. A bag contains 6 blue balls and 4 red balls. Two balls will be drawn from the bag. Let event A be the event that the first ball is blue, and let event B be the event that the second ball is blue. Then, the event A' will be the event that the first ball is red, and event B' will be the event that the second ball is red.

Since there are 6 blue balls out of a total of 10 balls, the probability of choosing a blue ball in the first drawing is 6/10. If a blue ball is taken out, then there will only be 5 blue balls and 9 total balls left; the probability of choosing a blue ball will be 5/9. On the other hand, if the first ball is a red ball, then there will be 6 blue balls and a total of 9 balls, in which case there would be a 6/9 (or 2/3) probability of getting a blue ball.

Therefore

P(A) = 6/10
P(A') = 1 - 6/10 = 4/10
P(B|A) = 5/9
P(B|A') = 2/3
P(A Ç B) = P(A) P(B|A) = (6/10)(5/9) = 1/3
P(A' Ç B) = P(A') P(B|A') = (4/10)(2/3) = 4/15

Since A and A' are mutually exclusive events, (B Ç A) and (B Ç A') are also mutually exclusive events. Thus, we can calculate P(B) as follows:

P(B)= P(B Ç S) = P( B Ç (A È A') ) = P( (B Ç A) È (B Ç A') )

= P(B Ç A) + P(B Ç A')
= 1/3 + 4/15
= 9/15

P (A È B) = P(A) + P(B) - P(A Ç B)

= 6/10 + 9/15 - 1/3
= 13/15

(A È B) is the event that either one of the two balls drawn is blue. This being the case, P(A È B) is the probability that the first ball is blue, plus the probability that the first ball is red and the second ball is blue. Thus,

P (A È B) = P(A) + P(A' Ç B)

= 6/10 + 4/15
= 13/15

A different way to obtain the solution, but the result is the same nevertheless.