7-3  Part I: Conditional Probability, Intersection of Events

Conditional probability

Experiment: I flip a coin twice
E1 = two heads = {HH}
P(E1) = ¼

BUT…

suppose I tell you that the outcome was the same for both flips, that is, that event

E2 = {HH, TT} has occurred

Question: what is the probability that E1 occurred GIVEN THAT (you already know that) E2 has occurred?
Answer:   (can you think what it would be?) ··

This is called a conditional probability.
Here's the mathematical notation:

P(E1 GIVEN E2)
or
P(E1 |  E2)

We can investigate conditional probability using (you guessed it!): ··:

• since E2 is given
• our sample space is now "restricted to" E2 = {TT, HH}
• in that restricted sample space
• there  is 1 chance out of 2 of E1 occurring
• and thus the probability we are looking for is:
P(E1 | E2) = 1/2

The KEY to conditional probability

 think of the given event  as  the whole sample space and compute probabilities  in  that restricted space

P(E1 GIVEN E2) = P(E1 | E2) =  = ··/··

But we can do it this way:  P(E1 GIVEN E2) = P(E1 | E2) =

Rewriting the last fraction above (divide numerator and denominator by n(S)):

So we have the formula:

P(A GIVEN B)
 For any two events A and B P(A | B) =

Again, we have the probability of a compound event in terms of the probabilities of set combinations of its simple events.

So for our specific example, P(E1 | E2) =  = 1/2, same answer as before!

These "clever" mathematical ways of formulating probabilities of compound events in terms of the probabilities of the constituent events constitute important mathematical tools for use in the study and calculation of probabilities.

The AND (intersection) of two events

Random experiment: flip coin twice
Event E1: get head on first throw = {HH, HT}
Event E2: get the same on both throws = {HH, TT}

Question: What is the probability of event E3 = E1 AND E2
that is, getting a head on first throw AND getting the same on the two throws?

Way 1:  compute the event E3 from its definition.  E3 = ··

P(E3) = ··
Way 2:      Use set theory

Given events A and B, the AND of those two events
is simply another event consisting of
all outcomes that are in both A and B

Definition of A AND B

 A AND B = A  B

Here goes:

We already know: P(E2 | E1) =  So (with a little algebra):
P(E1 Ç E2) = P(E1)P(E2 | E1) = (1/2)(1/2) = ¼

And we have the formula:

P(A AND B)
 For any two events A and B P(A  B) = P(A)P(B | A) = P(B)P(A | B)

Probability tables

Probabilities for various situations are frequently displayed in a standard tabular form, as follows.  Such tables often start as tables of counts, like the following classification of 200 adults according to gender and educational attainment:

 Education Male Female Totals Elementary 22 39 61 Secondary 44 56 100 College 22 17 39 Totals 88 112 200

This can be converted to an empirical probability table by dividing by 200:

 Education Male Female Totals Elementary .11 .195 .305 Secondary .22 .28 .5 College .11 .085 .195 Totals .44 .56 1

Vocabulary:
The green cells are called marginal probabilities (because they are shown in the margins of the table).
The blue cells are called joint probabilities, another name for the probability of the AND of two events.
The marginal probabilities represent certain totals of joint probabilities, as shown.

Now we can compute some probabilities:

P(college) = ··
P(college | female) = P(college AND female)/P(female) = ·· / ·· = ··
P(college | male) = P(college AND male)/P(male) = .11/.44 = .25

Notice that a conditional probability is (a joint probability) ÷ (a marginal probability)

Probability trees

A box contains 3 blue and 2 white balls. Two balls are drawn in succession, without replacement. Find the probability of drawing a white ball on the second draw.

Event notation:

B1 = Blue on 1st draw
W2 = White on 2nd draw, etc.
Method:  draw a tree that completely analyzes the problem:

P(W2) = 1/10 + 3/10 = 4/10 = 2/5

P(B2) = ··   +  ··   = ··   = 3/5 (naturally!)