Sum and Product Rules
Exercise. Consider tossing a coin ﬁve times. What is the proba-
bility of getting the same result on the ﬁrst two tosses or the last
Let E be the event that the ﬁrst two tosses are the same and F
be the event that the last two tosses are the same. Let n(E) be
the number of ways event E can occur.
Q. Is there any overlap in E and F? I.e., is E \ F = ;?
n(E and F) = n(E \ F)
1 Now we can calculate the probability P(E or F):
P(E or F) =
Theorem (The Sum Rule) If E and F are events in an experi-
ment then the probability that E or F occurs is given by:
P(E or F) = P(E) + P(F) ▯ P(E and F)
Example. What is the probability when a pair of dice are rolled
that at least one die shows a 5 or the dice sum to 8?
2 Exercise. Given a bag of 3 red marbles, 5 black marbles and 8
green marbles select one marble and then a second. What is the
probability that both are red?
Q. What is the probability of the ﬁrst marble being red?
Q. What is the probability of the second marble being red?
Q. Probability of both marbles being red?
When the probability of an event E depends on a previous event
F happening we denote this probability as P(EjF).
3 Theorem (The Product Rule). If E and F are two events in an
experiment then the probability that both E and F occur is:
(▯) P(E and F) = P(F) ▯ P(EjF) = P(E) ▯ P(FjE)
Q. What does it mean if P(EjF) = P(E)?
Q. Therefore, if E and F are two independent events, what is
Example. Suppose there is a noisy communication channel in
which either a 0 or a 1 is sent with the following probabilities:
▯ Probability a 0 is sent is 0.4.
▯ Probability a 1 is sent is 0.6.
▯ Probability that due to noise, a 0 is changed to a 1 during
transmission is 0.2.
▯ Probability that due to noise, a 1 is changed to a 0 during
transmission is 0.1.
Suppose that a 1 is received. What is the probability that a 1 was
4 Let A denote that a 1 was received and B denote the event that
a 1 was sent.
Q. What is the probability that we are solving for?
How can we solve for this? The Product Rule says that:
P(A and B) = P(B) ▯ P(AjB) = P(A)P(BjA)