If you roll a six-sided die, there are six possible outcomes,
and each of these outcomes is equally likely. A six is as
likely to come up as a three, and likewise for the other four
sides of the die. What, then, is the probability that a one
will come up? Since there are six possible outcomes, the
probability is 1/6
16. What is the probability that either a one or a six
will come up? The two outcomes about which we are concerned (a
one or a six coming up) are called favorable outcomes. Given
that all outcomes are equally likely, we can compute the
probability of a one or a six using the formula:
probability=Number of favorable outcomesNumber of possible equally-likely outcomes
probability
Number of favorable outcomes
Number of possible equally-likely outcomes
In this case there are two favorable outcomes and six possible
outcomes. So the probability of throwing either a one or six
is 1/3
13. Don't be misled by our use of the term
favorable, by the way. You should understand it
in the sense of "favorable to the event in question
happening." That event might not be favorable to your
well-being. You might be betting on a three, for example.
The above formula applies to many games of chance. For
example, what is the probability that a card drawn at random
from deck of playing cards will be an ace? Since the deck has
four aces, there are four favorable outcomes; since the deck
has 52 cards, there are 52 possible outcomes. The probability
is therefore
4/52=1/13
452
113
. What about the probability that the
card will be a club? Since there are 13 clubs, the probability
is
13/52=1/4
1352
14
.
Let's say you have a bag with 20 cherries, 14 sweet and 6
sour. If you pick a cherry at random, what is the probability
that it will be sweet? There are 20 possible cherries that
could be picked, so the number of possible outcomes is 20. Of
these 20 possible outcomes, 14 are favorable (sweet), so the
probability that the cherry will be sweet is
14/20=7/10
1420
710
. There is one potential complication to this
example, however. It must be assumed that the probability of
picking any of the cherries is the same as the probability of
picking any other. This wouldn't be true if (let us imagine)
the sweet cherries are smaller than the sour ones. (The sour
cherries would come to hand more readily when you sampled from
the bag.) Let us keep in mind, therefore, that when we assess
probabilities in terms of the ratio of favorable to all
potential cases, we rely heavily on the assumption of equal
probability for all outcomes.
Here is a more complex example. You throw 2 dice. What is the
probability that the sum of the two dice will be 6? To solve
this problem, list all the possible outcomes. There are 36 of
them since each die can come up one of six ways. The 36
possibilities are shown below.
|
Die 1
|
Die 2
|
Total
|
Die 1
|
Die 2
|
Total
|
Die 1
|
Die 2
|
Total
|
|
1
|
1
|
2
|
3
|
1
|
4
|
5
|
1
|
66
|
|
1
|
2
|
3
|
3
|
2
|
5
|
5
|
2
|
7
|
|
1
|
3
|
4
|
3
|
3
|
6
6
|
5
|
3
|
8
|
|
1
|
4
|
5
|
3
|
4
|
7
|
5
|
4
|
9
|
|
1
|
5
|
6
6
|
3
|
5
|
8
|
5
|
5
|
10
|
|
1
|
6
|
7
|
3
|
6
|
9
|
5
|
6
|
11
|
|
2
|
1
|
3
|
4
|
1
|
5
|
6
|
1
|
7
|
|
2
|
2
|
4
|
4
|
2
|
6
6
|
6
|
2
|
8
|
|
2
|
3
|
5
|
4
|
3
|
7
|
6
|
3
|
9
|
|
2
|
4
|
6
6
|
4
|
4
|
8
|
6
|
4
|
10
|
|
2
|
5
|
7
|
4
|
5
|
9
|
6
|
5
|
11
|
|
2
|
6
|
8
|
4
|
6
|
10
|
6
|
6
|
12
|
You can see that 5 of the 36 possibilities total 6. Therefore,
the probability is
5/36
536.
If you know the probability of an event
occurring, it is easy to compute the probability that the event
does not occur. If
PrA
A
is the probability of Event A, then
1-PrA
1
A
is the probability that the event does not occur. For the
last example, the probability that the total is 6 is
5/36
536. Therefore, the probability that the total is not 6
is
1-5/36=31/36
1
536
3136
.
Events A and B are independent events
if the probability of Event B occuring is the same whether or
not Event A occurs. Let's take a simple example. A fair coin
is tossed two times. The probability that a head comes up on
the second toss is 1/2 12 regardelss of
whether or not a head came up on the first toss. The two
events are
first toss is a head, ; and second toss is
a head. So these events are
independent. Consider the two events
"It will rain tomorrow in Houston"; and
"It will rain tomorrow in Galveston (a city near
Houston)". These events are not independent
because it is more likely that it will rain in Galveston on
days it rains in Houston than on days it does not.
When two events are independent, the probability of both
occuring is the product of the probabilities of the
individual events. More formally, if events A and B are
independent, then the probability of both A and B occuring
is:
PrA∧B=PrAPrB
A
B
A
B
where
PrA∧B
A
B
is the probability of events A and B both occuring,
PrA
A
is the probability of event A occuring, and
PrB
B
is the probability of event B occuring.
If you flip a coin twice, what is the probability that it
will come up heads both times? Event A is that the coin
comes up heads on the first flip and Event B is that the
coin comes up heads on the second flip. Since both
PrA
A
and
PrB
B
equal
1/2
12, the probability that both events occur is
1/2×1/2=1/4
12
12
14
Lets take another example. If you flip a coin and
roll a six-sided die, what is the probability that the coin
comes up heads and the die comes up 1? Since the two events
are independent, the probability is simply the probability
of a head (which is
1/2
12) times the probability of the die
coming up 1 (which is
1/6
16). Therefore, the probability of
both events occuring is
1/2×1/6=1/12
12
16
112
.
One final example: You draw a card from a deck of cards, put
it back, and then draw another card. What is the probability
that the first card is a heart and the second card is black?
Since there are 52 cards in a deck, and 13 of them are
hearts, the probability that the first card is a heart is
13/52=1/4
1352
14
. Since there are 26 black cards in the deck, the
probability that the second card is black is
26/52=1/2
2652
12
. The probability of both events occurring is therefore
1/4×1/2=1/8
14
12
18
.
If Events A and B are independent, the probability that
either Event A or Event B occurs is:
PrA∨B=PrA+PrB-PrA∧B
A
B
A
B
A
B
In this discussion, when we say "A or B occurs" we include
three possibilities:
- A occurs and B does not occur
- B occurs and A does not ocuur
- Both A and B occur
This use of the word "or" is technically called
inclusive or because it includes the case in
which both A and B occur. If we included only the first two
cases, then we would be using an
exclusive or.
(Optional) We can derive the law for
PrA∨B
A
B
from our law about
PrA∧B
A
B
. The event
A∨B
A
B
can happen in any of the following ways:
-
A∧B
A
B
happens
-
A∧¬B
A
B
happens
-
¬A∧B
A
B
happens.
The simple event A can happen if either
A∧B
A
B
happens, or
A∧¬B
A
B
happens. Similarly, the simple event B
happens if either
A∧B
AB
happens or
¬A∧B
A
B
happens.
PrA+PrB
A
B
is therefore
PrA∧B+PrA∧¬B+PrA∧B+Pr¬A∧B
A
B
A
B
A
B
A
B
whereas
PrA∨B
A
B
is
PrA∧B+PrA∧¬B+Pr¬A∧B
A
B
A
B
A
B
. We can make these two sums equal by subtracting one
occurrence of
PrA∧B
A
B
from the first. Hence,
PrA∨B=PrA+PrB-PrA∧B
A
B
A
B
A
B
.
Now for some examples. If you flip a coin two times, what is
the probability that you will get a head on the first flip or
a head on the second flip (or both)? Letting Event A be a head
on the first flip and Event B be head on the second flip then
PrA=1/2
A
12
,
PrB=1/2
B
12
, and
=PrA∧B1/4
A
B
14
. Therefore,
PrA∨B=1/2+1/2-1/4
A
B
12
12
14
. If you throw a six-sided die and then flip a coin,
what is the probability that you will get either a 6 on the
die or a head on the coin flip (or both)? Using the formula,
Pr6∨head=Pr6++Prhead+Pr6∧head=1/6+1/2-1/6×1/2=7/12
6
head
6
head
6
head
16
12
16
12
712
An alternate approach to computing this value is to start by
computing the probability of not getting either a 6 or a
head. Then subtract this value from 1 to compute the
probability of getting a 6 or a head. Although this is a
complicated method, it has the advantage of being applicable
to problems with more than two events. Here is the calculation
in the present case. The probability of not getting either a 6
or a head can be recast as the probability of
(not getting a 6)∧(not getting a head)
(not getting a 6)
(not getting a head)
This follows because if you did not get a 6 and you did not
get a head, then you did not get a 6 or a head. The
probability of not getting a six is
1-1/6=5/6
1
16
56
. The probability of not getting a head is
1-1/2=1/2
1
12
12
. The probability of not getting a six and not getting a
head is
5/6×1/2=5/12
56
12
512
. This is therefore the probability of
not getting a 6 or a head. The probability of getting a six or
a head is therefore (once again)
1-5/12=7/12
1
512
712
.
If you throw a die three times, what is the probability that
one or more of your throws will come up with a 1? That is,
what is the probability of getting a 1 on the first throw OR a
1 on the second throw OR a 1 on the third throw? The easiest
way to approach this problem is to compute the probability of
- NOT getting a 1 on the first throw
- AND not getting a 1 on the second throw
- AND not getting a 1 on the third throw.
The answer will be 1 minus this probability. The probability
of not getting a 1 on any of the three throws is
5/6×5/6×5/6=125/216
56
56
56
125216
. Therefore, the probability of getting a 1 on at
least one of the throws is
1-125/216=91/216
1
125216
91216
.
Often it is required to compute the probability of an event
given that another event has occured. For example, what is the
probability that two cards drawn at random from a deck of
playing cards will both be aces? It might seem that you could
use the formula for the probability of two independent events
and simply multiply
4/52×4/52=1/169
452
452
1169
. This would be
incorrect, however, because the two events are not
independent. If the first card drawn is an ace, then the
probability that the second card is also an ace would be lower
because there would only be three aces left in the deck.
Once the first card chosen is an ace, the probability that the
second card chosen is also an ace is called the conditional probability
of drawing an ace. In this case the condition is
that the first card is an ace. Symbolically, we write this as:
Prace on second draw|an ace on the first draw
an ace on the first draw
ace on second draw
The vertical bar "|" is read as "given," so the above
expression is short for "The probability that an ace is drawn on the
second draw given that an ace was drawn on the first draw." What is
this probability? Since after an ace is drawn on the first draw, there
are 3 aces out of 51 total cards left. This means that the probability
that one of these aces will be drawn is
3/51=1/17
351
117
.
If Events A and B are not independent, then
PrA∧B=PrAPrB|A
A
B
A
A
B
.
Applying this to the problem of two aces, the probability of drawing
two aces from a deck is
4/52×3/51=1/221
452
351
1221
.
One more example: If you draw two cards from a deck, what is
the probability that you will get the Ace of Diamonds and a
black card? There are two ways you can satisfy this condition:
(a) You can get the Ace of Diamonds first and then a black
card or (b) you can get a black card first and then the ace of
diamonds. Let's calculate Case A. The probability that the
first card is the Ace of Diamonds is
1/52
152. The probability that the second card is black given
that the first card is the Ace of Diamonds is
26/51
2651 because 26 of the remaining 51 cards are black. The
probability is therefore
1/52×26/51=1/102
152
2651
1102
. Now for Case B: the probability that the first card
is black is
26/52=1/2
2652
12
. The probability that the second card is the Ace of
Diamonds given that the first card is black is
1/51
151. The
probability of Case 2 is therefore
1/2×1/51=1/102
12
151
1102
, the same as the probability of Case 1. Recall that
the probability of A or B is
=PrA+PrB-PrA∧B
A
B
A
B
. In this problem,
PrA∧B=0
A
B
0
since a card cannot be the Ace of Diamonds and be a
black card. Therefore, the probability of Case A or Case B is
1/101+1/101=2/101
1101
1101
2101
. So,
2/101
2101 is the probability that you will get the Ace of
Diamonds and a black card when drawing two cards from a deck.
If there are 25 people in a room, what is the probability that
at least two of them share the same birthday. If your first
thought is that it is
25/365=0.068
25365
0.068
, you will be surprised to learn it is much higher
than that. This problem requires the application of the
sections on
PrA∧B
A
B
and conditional probability.
This problem is best approached by asking what is the
probability that no two people have the same birthday. Once we
know this probability, we can simply subtract it from 1 to
find the probability that two people share a birthday.
If we choose two people at random, what is the probability
that they do not share a birthday? Of the 365 days on which
the second person could have a birthday, 364 of them are
different from the first person's birthday. Therefore the
probability is
364/365
364365. Let's define
Pr
2
Pr
2
as the probability
that the second person drawn does not share a birthday with
the person drawn previously.
Pr
2
Pr
2
is therfore
364/365
364365. Now define
Pr
3
Pr
3
as the probability that the third
person drawn does not share a birthday with anyone drawn
previously given that there are no
previous birthday matches.
Pr
3
Pr
3
is therefore a conditional
probability. If there are no previous birthday matches, then
two of the 365 days have been "used up," leaving 363
non-matching days. Therefore
Pr
3
=363/365
Pr
3
363365
. In like manner,
Pr
4
=362/365
Pr
4
362365
,
Pr
5
=361/365
Pr
5
361365
, and so on up to
Pr
25
=341/365
Pr
25
341365
.
In order for there to be no matches, the second person must
not match any previous person and the third person must not
match any previous person, and the fourth person must not
match any previous person, etc. Since
PrA∧B=PrAPrB
A
B
A
B
,
all we have to do is multiply
Pr
2
Pr
2
,
Pr
3
Pr
3
,
Pr
4
Pr
4
...
Pr
25
Pr
25
together. The
result is 0.431. Therefore the probability of at least one
match is 0.561.
A fair coin is flipped five times and comes up heads each
time. What is the probability that it will come up heads on
the sixth flip? The correct answer is, of course,
1/2 12. But many
people believe that a tail is more likely to occur after
throwing five heads. Their faulty
reasoning may go something like this "In the
long run, the number of heads and tails will be the same, so
the tails have some catching up to do". The flaws in
this logic are exposed in the simulation in the next section.
- favorable outcome:
A favorable outcome is the outcome of interest. For example
one could define a favorable outcome in the flip of a coin as
a head. The term favorable outcome does not
necessarily mean that the outcome is desirable - in some
experiments, the favorable outcome could be the failure of a
test, or the occurrence of an undesirable event.
- independent events:
Intuitively, two events A and B are independent if the
occurrence of one has no effect on the probability of the
occurrence of the other. For example, if you throw two dice,
the probability that the second one comes up 1 is independent
of whether the first die came up 1. Formally, this can be
stated in terms of
conditional
probabilities:
PrA|B=PrA
B
A
A
and
PrB|A=PrB
A
B
B
.
- conditional probability:
The probability that event A occurs given that event B has
already occurred is called the conditional probability of A
given B. Symbolically, this is written as
PrA|B
B
A
. The probability it rains on Monday given that it
rained on Sunday would be written as
PrRain on Monday|Rain on Sunday
Rain on Sunday
Rain on Monday
.
Introduction to Probability