David M. Lane
- 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
- Describe the gamblers' 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,
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.
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
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
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
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.
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.
Please answer the questions: