CHAPTER 5
PROBABILITY
Probability theory allows us to make some sense out of happenings due to chance. It is quite remarkable that mathematical rules can apply to random events.
5A Probability Rules
If you flip a coin many times, about half the time you get heads and the other half you get tails. In general, as you flip the coin more and more, the ratio of heads to tails comes closer and closer to one. Why should this always be so? Is there some “god of probability” sitting somewhere counting the heads and tails, and perhaps fudging the coin a little so that the ratio of heads to tails does not stray too far from one? Regardless of the reason, there is a mathematical rule governing coin flipping  it says that when you flip a coin the outcomes are about even between heads and tails.
In flipping a coin there are two possible “events”. We label as “H” the event of getting a head, and as “T” the event of getting a tail. Since each event happens half the time over the long haul, we say that the probability of H is 1/2 and the probability of T is 1/2. The corresponding equations are
P(H) = 1/2 , P(T) = 1/2 .
Remember, the probabilities refer to what happens if you flip the coin many times. They do not say, for example, that if you flip a coin twice you will get one head and one tail, or that if you flip the coin 10 times you will get exactly 5 heads and 5 tails. What they do say is that as you flip the coin more and more, the ratio of heads to tails will become closer to one.
The above assignment of probabilities is due largely to experience  after much coin flipping we observe that heads occur about half the time and tails about the same. However, we can also analyze the coin and try to guess what it is about the coin that makes the two probabilities equal. One of the first things you notice about a coin is its symmetry  there is no difference between the two sides except for the engravings. Thus we see no reason why one side should have an advantage over the other in landing face up. Therefore, in the absence of any compelling reason to believe otherwise, we would expect that heads and tails appear with equal frequency. We say that the events heads and tails are equally likely.
We consider now a situation where events are not equally likely. Suppose that a certain basketball player hits 75%  or 3/4  of her free throws. This does not mean that every time she shoots 4 free throws she will hit exactly 3  it means rather that if she shoots a very large number of free throws, say over the course of a whole season, she will hit about 75% of them. When she shoots any one particular free throw there are two possible events  she can hit the shot (denoted as H) with probability 3/4, or she can miss it (denoted as M) with probability 1/4. These probabilities may be written as
P(H) = 3/4 , P(M) = 1/4 .
Note that the two possible events in this example can be viewed as negations of one another  that is, if the event “hits” occurs, then the event “misses” does not occur, while if “hits” does not occur then “misses” does occur. The same can be said of the coin flipping problem  there are two events, “heads” and “tails”, and one and only one of the two events must occur. In general, two events E and F are said to be complementary, or negations of one another, if
1) whenever event E occurs, event F cannot occur, 
2) whenever event E does not occur, event F must occur. 
When events E and F are complementary we signify this by writing
E = ~ F , F = ~ E.
Thus, in the free throw example we have H = ~ M and M = ~ H, while in the coin flipping example, H = ~ T and T = ~ H.

In probability theory, an experiment is some activity in which the outcome is due to chance. Thus, flipping a coin can be viewed as an experiment, and so can shooting a free throw. Another experiment, popular in various board and gambling games, consists of throwing a single die. In this experiment there are six possible outcomes, which we label with the numbers 1, 2, 3, 4, 5, 6, corresponding to the numbers of dots on the six sides of the die. For example, we say that the event “4” occurs if the side with four dots lands facing upwards, as on the die shown. Because a die is symmetric in the six sides, there is no reason to believe that one side is favored over any other, and so we would guess that all of the six possible outcomes occur with equal frequency. Real life experience with throwing dice bears out this presumption  that is, if a die is thrown a great number of times, then each of the six outcomes will occur about 1/6 of the time, and this description becomes more and more accurate as the number of throws increases. The probabilities of the six possible outcomes may be displayed with the chain of equations
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6 .
But in throwing a die there are other possible events besides these six elementary outcomes listed. For example, we might consider also the events
E : an even number is thrown , F : a number divisible by 3 is thrown .
What is the probability of the event E? The number facing up on the die is even if it is any of the three numbers 2, 4, or 6. The number 2, when the die is thrown many times, will appear 1/6 of the time, as will also the numbers 4 and 6. Hence the proportion of times the number thrown will be even is the sum 1/6 + 1/6 + 1/6 = 3/6 = 1/2, and consequently the probability of E is P(E) = 1/2. Likewise, the event F occurs if the number facing up is 3 or 6, so the probability of event F is P(F) = 1/6 + 1/6 = 2/6 = 1/3.
Two other events we might describe in throwing a die are
G : the number thrown is positive , H : the number thrown is negative .
The event G occurs if either 1, 2, 3, 4, 5, or 6 is thrown; thus its probability is
P(G) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1 .
This event is called a certain event  it must always occur. In contrast, event H is an impossible event, in that it can never occur. Its probability is P(H) = 0/6 = 0.
We are now ready to discuss the general rules of probability. The probability of an event is a number expressing the proportion of times the event occurs when an experiment is performed many times. If the event is impossible then it never occurs, so its probability is 0. On the other hand, if the event is certain then it must occur 100% of the times, so its probability is 1. All other events, neither impossible nor certain, occur at a proportion somewhere between 0 and 1. These observations explain our first three rules:
Rule 1 
An impossible event has probability 0. 
Rule 2 
A certain event has probability 1. 
Rule 3 
For any event E, 0 ≤ P(E) ≤ 1. 
Why so many rules ? 
The fourth rule concerns complementary events. If a basketball player hits 3/4 of her free throws, then she misses 1 − 3/4 = 1/4 of them. Likewise, if a coin lands heads 1/2 the time, then it lands tails 1 − 1/2 = 1/2 of the time. In general, if the proportion of times an event occurs is p, then the proportion of times its negation, or complementary event, occurs is 1 − p; i.e., the probability of an event and the probability of its complementary event must add to 1:
Rule 4 
For any event E, P(E) + P(~E) = 1. 
Finally we discuss Rule 5. Going back to throwing one die, let us consider the events
E : an even number is thrown, 
F : a number divisible by 3 is thrown, 
E Ù F : a number both even and divisible by 3 is thrown, 
E Ú F : a number either even or divisible by 3 is thrown. 
(We use the logic symbols “Ù” for “and”, and “Ú” for “or”.) As we have seen, E occurs when the number on the die is 2, 4, or 6, so P(E) = 3/6 = 1/2, while F occurs if the number is 3 or 6, so P(F) = 2/6 = 1/3. The event E Ù F occurs only when the number on the die is 6, so P(E Ù F) = 1/6. The event E Ú F is a little more tricky. It is a common mistake to add P(E) and P(F) to get the wrong answer,
P(E Ú F) = P(E) + P(F) = 3/6 + 2/6 = 5/6 ,
as the probability of E Ú F. However, when you add 3/6 to 2/6 you are adding the contribution of the number 6 twice, because as an outcome of E Ù F the number 6 contributes to event E as well as to event F. Thus we must subtract this contribution of E Ù F, getting
P(E Ú F) = P(E) + P(F) − P(E Ù F) = 3/6 + 2/6 − 1/6 = 4/6 = 2/3 .
This example illustrates our last rule of probability,
Rule 5 
For events E and F, P(E Ú F) = P(E) + P(F) − P(E Ù F). 
example 1
Suppose the weatherman estimates the probability of rain tomorrow as 60%, the probability of lightning as 50%, and the probability of both rain and lightning as 20%. We determine the probability that tomorrow there will be
a) no rain,
b) no lightning,
c) rain or lightning,
d) rain but no lightning,
e) lightning but no rain,
f) neither rain nor lightning.
We let R and L denote the events
R : it rains tomorrow , L : there is lightning tomorrow.
Changing the percentages to fractions, we may write
P(R) = 60% = 6/10 , P(L) = 50% = 5/10 , P(R Ù L) = 20% = 2/10 .
The negation of event R is the event of no rain, while the negation of L is the event of no lightning. By rule 4, the probabilities of a) no rain and b) no lightning are, respectively,
P(~R) = 1 − P(R) = 1 − 6/10 = 4/10 = 40%, 
P(~L) = 1 − P(L) = 1 − 5/10 = 5/10 = 50%. 
Thus there is a 40% probability of no rain, and a 50% probability of no lightning.
The probability of rain or lightning is given by Rule 5 as
P(R Ú L) = P(R) + P(L) − P(R Ù L) = 60% + 50% − 20% = 90% .
(The probability of R Ù L, counted twice in the sum P(R) + P(L), must be subtracted.)
In solving parts d), e), and f), it is perhaps easiest to resort to a Venn diagram. When the weatherman says the probability of rain tomorrow is 60%, he means that, among days having conditions similar to those of today, about 60% of them have rain the next day. The universal set in our Venn diagram represents all days having conditions similar to those of today. We draw circles R and L representing days followed by rain and days followed by lightning, respectively. We write 2/10 in the intersection of R and L, representing P(R Ù L), the proportion of days followed by both rain and lightning. Since the R circle must contain 6/10 of all the days, we write 4/10 in the region inside the R circle but outside the L circle. Similarly, we write 3/10 in the region inside the L circle but outside the R circle, because the L circle must contain 5/10 of the days. Finally, the proportions in all the regions must add to 1, since the universal rectangle contains 100% of the days; thus outside the two circles we write the fraction
1 − (4/10 + 2/10 + 3/10) = 1 − 9/10 = 1/10 .
We can now read the answers to d), e), and f) from the Venn diagram. The probability of rain but no lightning is 4/10 = 2/5, the probability of lightning but no rain is 3/10, and the probability of neither rain nor lightning is 1/10.
In a probability experiment, two events E and F are said to be mutually exclusive if they cannot both occur. In such a case the event E Ù F is impossible, making P(E Ù F) = 0. Then in Rule 5 the formula for P(E Ú F) simplifies to
P(E Ú F) = P(E) + P(F) .
For instance, when we throw a die the events
E : an even number is thrown , O : an odd number is thrown
are mutually exclusive because not both can occur. Therefore,
P(E Ú O) = P(E) + P(O) = 3/6 + 3/6 = 1 .
On the other hand, the events
E : an even number is thrown , F : a number divisible by 3 is thrown
are not mutually exclusive, as both events occur if the number thrown is 6. In this case,
P(E Ú F) ≠ P(E) + P(F) ,
because the contribution of the outcome 6 is counted double in the sum on the right.
Analogous comments hold for more than two events. For example, if no two of the events E, F, and G can occur at the same time, then
P(E Ú F Ú G) = P(E) + P(F) + P(G) ,
as no double counting occurs on the right. An example in the throwing of a die are the three events
2 : a 2 is thrown , 4 : a 4 is thrown , 6 : a 6 is thrown .
As only one of these can occur on the same throw, the probability of an even number is
P(even number) = P(2 Ú 4 Ú 6) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2 .
example 2
The three candidates for homecoming queen are Agnes, Bea, and Cyd. Madam Trafalger, the beauty expert, estimates that the probability Agnes will win is P(A) = 1/2, and the probability Bea will win is P(B) = 1/3. We will compute the probability that
a) Agnes or Bea will win,
b) Cyd will win,
c) Agnes or Bea will not win,
d) Dot will win.
a) As not both Agnes and Bea can win, these events are mutually exclusive; thus
P(A Ú B) = P(A) + P(B) = 1/2 + 1/3 = 5/6 .
b) The event C that Cyd wins is the negation of the event A Ú B, when Agnes or Bea wins; therefore,
P(C) = 1 − P(~C) = 1 − P(A Ú B) = 1 − 5/6 = 1/6 .
c) The event that Agnes or Bea will not win is a certain event, as obviously not both of them can win; thus the probability that Agnes or Bea will not win is 1.
d) Dot cannot win because she is not even in the contest; thus the probability she will win is 0.
EXERCISES 5A
R : a red card is drawn, 
B : a black card is drawn, 

H : a heart is drawn, 
C : a club is drawn, 

F : a face card is drawn, 
N : a numerical card is drawn, 

K : a king is drawn, 
E : an eight is drawn. 
a. an even number? 

b. a number divisible by 3? 

c. an odd number or a number divisible by 3? 

d. a number both even and divisible by 3? 

e. a number larger than 7? 

f. a number larger than 7 or less than 4? 

g. a number larger than 7 and less than 4? 

h. a number larger than 7 and larger than 4? 

i. a number less than 7 or larger than 4? 
a. both rain and hail? 


e. no rain? 

b. neither rain nor hail? 
f. no hail?  
c. rain but no hail? 
g. either rain or hail, but not both? 

d. hail but no rain? 
h. not both rain and hail? 
a. a female win? 

b. a male win? 

c. Angie, Beth, or Chuck win? 

d. Dan win? 

e. Beth not win? 

f. Beth or Chuck not win? 

g. Angie and Chuck not win? 

h. Edith win? 
a. miss her math class? 

b. attend both her classes? 

c. miss both her classes? 

d. attend math class but miss French class? 

e. attend French class but miss math class? 

f. attend one class but miss the other? 
a. an even number? 

b. a number divisible by 3? 

c. a number less than 4? 