INTRODUCTION TO PROBABILITY.
BASI CONCEPTS.
Learning Objectives
- Compute probability in a situation where there are equally-likely outcomes
- Apply concepts to cards and dice
- Compute the probability of two independent events both occurring
- Compute the probability of either of two independent events occurring
- Do problems that involve conditional probabilities
- Compute the probability that in a room of N people, at least two share
a birthday
- Describe the gambler's fallacy
Probability of a Single Event
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. 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:
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. 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 a 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. What about the probability that the card will be
a club? Since there are 13 clubs, the probability is 13/52 = 1/4.
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. 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
|
6
|
|
1
|
2
|
3
|
|
3
|
2
|
5
|
|
5
|
2
|
7
|
|
1
|
3
|
4
|
|
3
|
3
|
6
|
|
5
|
3
|
8
|
|
1
|
4
|
5
|
|
3
|
4
|
7
|
|
5
|
4
|
9
|
|
1
|
5
|
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
|
2
|
8
|
|
2
|
3
|
5
|
|
4
|
3
|
7
|
|
6
|
3
|
9
|
|
2
|
4
|
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.
If you know the probability of an event occurring,
it is easy to compute the probability that the event does not
occur. If P(A) is the probability of Event A, then 1 - P(A) is
the probability that the event does not occur. For the last example,
the probability that the total is 6 is 5/36. Therefore, the probability
that the total is not 6 is 1 - 5/36 = 31/36.
Probability of Two (or more) Independent Events
Events A and B are independent
events if the probability of Event B occurring 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 regardless of whether
or not a head came up on the first toss. The two events are
(1) first toss is a head and (2) second toss is a head. So
these events are independent. Consider the two events (1) "It
will rain tomorrow in Houston"
and (2) "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.
Probability of A and B
When two events are independent, the probability
of both occurring 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 occurring is:
P(A and B) = P(A) x P(B)
where P(A and B) is the probability of events
A and B both occurring, P(A) is the probability of event A occurring,
and P(B) is the probability of event B occurring.
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 P(A) and P(B) equal
1/2, the probability that both events occur is
1/2 x 1/2 = 1/4
Let's 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) times the probability of the die coming
up 1 (which is 1/6). Therefore, the probability of both events
occurring is 1/2 x 1/6 = 1/12.
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. Since there are
26 black cards in the deck, the probability that the second card
is black is 26/52 = 1/2. The probability of both events occurring
is therefore 1/4 x 1/2 = 1/8.
Probability of A or B
If Events A and B are independent, the probability
that either Event A or Event B occurs is:
P(A or B) = P(A) + P(B) - P(A and 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 occur
- 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 P(A-or-B)
from our law about P(A-and-B). The event "A-or-B"
can happen in any of the following ways:
- A-and-B happens
- A-and-not-B happens
- not-A-and-B happens.
The simple event A can happen if either A-and-B
happens or A-and-not-B happens. Similarly, the simple event
B happens if either A-and-B happens or not-A-and-B happens.
P(A) + P(B) is therefore P(A-and-B) + P(A-and-not-B) + P(A-and-B)
+ P(not-A-and-B), whereas P(A-or-B) is P(A-and-B) + P(A-and-not-B)
+ P(not-A-and-B). We can make these two sums equal by subtracting
one occurrence of P(A-and-B) from the first. Hence, P(A-or-B)
= P(A) + P(B) - P(A-and-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 a head on the second flip,
then P(A) = 1/2, P(B) = 1/2, and P(A and B) = 1/4. Therefore,
P(A or B) = 1/2 + 1/2 - 1/4 = 3/4.
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,
P(6 or head) = P(6) + P(head) - P(6 and head)
=
(1/6) + (1/2) - (1/6)(1/2)
=
7/12
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) AND (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. The probability
of not getting a head is 1 - 1/2 = 1/2. The probability of not
getting a six and not getting a head is 5/6 x 1/2 = 5/12. 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.
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
x 5/6 x 5/6 = 125/216. Therefore, the probability of getting a
1 on at least one of the throws is 1 - 125/216 = 91/216.
Conditional Probabilities
Often it is required to compute the probability
of an event given that another event has occurred. 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 x 4/52 = 1/169. 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:
P(ace on second draw | an ace on the first 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.
If Events A and B are not independent, then
P(A and B) = P(A) x P(B|A).
Applying this to the problem of two aces, the
probability of drawing two aces from a deck is 4/52 x 3/51 = 1/221.
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. The probability that the second card
is black given that the first card is the Ace of Diamonds is 26/51
because 26 of the remaining 51 cards are black. The probability
is therefore 1/52 x 26/51 = 1/102. Now for Case B: the probability
that the first card is black is 26/52 = 1/2. The probability that
the second card is the Ace of Diamonds given that the first card
is black is 1/51. The probability of Case B is therefore 1/2 x
1/51 = 1/102, the same as the probability of Case A. Recall that
the probability of A or B is P(A) + P(B) - P(A and B). In this
problem, P(A and 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 1/102 + 1/102 = 2/102 = 1/51. So, 1/51 is the probability that
you will get the Ace of Diamonds and a black card when drawing
two cards from a deck.
Birthday Problem
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, you will be
surprised to learn it is much higher than that. This problem requires
the application of the sections on P(A and 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. Let's define P2 as the probability that
the second person drawn does not share a birthday with the person
drawn previously. P2 is therefore 364/365. Now define P3 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. P3 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 P3 = 363/365. In like manner, P4
= 362/365, P5 = 361/365, and so on up to P25 = 341/365.
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
P(A and B) = P(A)P(B), all we have to do is multiply P2, P3, P4
...P25 together. The result is 0.431. Therefore the probability
of at least one match is 0.569.
Gambler's Fallacy
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.
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 this chapter.
Question 2 out of 12.
You have a bag of marbles. There are 3 red marbles, 2 green marbles, 7
yellow marbles, and 3 blue marbles. What is the probability of drawing a
yellow or red marble?
Figure out the Big Picture: Find
out why you are doing this math. How does it fit with your other
courses (science, geography, English, engineering)? You could do some
Internet searches on the math you are studying and include
"application". Get a sense of where you are going and why you are doing
this. Mathematics is compulsory in most of the world – there has to be a
reason…
Use more than one resource.
It often happens that you can’t follow the teacher’s explanation and
your textbook is very confusing. Borrow 2 or 3 textbooks similar to your
own from your library and read what they have to say about the topic.
Often they will have a diagram, a picture or an explanation that gives
you the "Ahhh – I get it!" that you desire.
Time Management.
Start homework assignments as soon as you get them. There may be some
things on there that you haven’t done in class yet (because maybe it is
not due for a few weeks). That’s good – it helps to focus your thoughts
so that when you are doing that section in class, you know that it is
important and you’ll know what you don’t know. Nobody plans to fail –
but many fail to plan…
Keep a sense of humour!
Don’t lose the ability to laugh at yourself and your own mistakes.
Mistakes are not the end of the world – they are the beginning of real
learning!