Page 471 - Discrete Mathematics and Its Applications

450 7 / Discrete Probability

Hence,
|E 1 ∪ E 2 |
p(E 1 ∪ E 2 ) =
|S|
|E 1 |+|E 2 |−|E 1 ∩ E 2 |
=
|S|
|E 1 | |E 2 | |E 1 ∩ E 2 |
= + −
|S| |S| |S|
= p(E 1 ) + p(E 2 ) − p(E 1 ∩ E 2 ).

EXAMPLE 9 What is the probability that a positive integer selected at random from the set of positive integers
not exceeding 100 is divisible by either 2 or 5?

Solution: Let E 1 be the event that the integer selected at random is divisible by 2, and let E 2 be
the event that it is divisible by 5. Then E 1 ∪ E 2 is the event that it is divisible by either 2 or 5.
Also, E 1 ∩ E 2 is the event that it is divisible by both 2 and 5, or equivalently, that it is divisible
by 10. Because |E 1 |= 50, |E 2 |= 20, and |E 1 ∩ E 2 |= 10, it follows that

p(E 1 ∪ E 2 ) = p(E 1 ) + p(E 2 ) − p(E 1 ∩ E 2 )
50 20 10 3 ▲
= + − = .
100 100 100 5

Probabilistic Reasoning

A common problem is determining which of two events is more likely. Analyzing the probabil-
ities of such events can be tricky. Example 10 describes a problem of this type. It discusses a
famous problem originating with the television game show Let’s Make a Deal and named after
the host of the show, Monty Hall.
EXAMPLE 10 The Monty Hall Three-Door Puzzle Suppose you are a game show contestant. You have a
chance to win a large prize. You are asked to select one of three doors to open; the large prize
is behind one of the three doors and the other two doors are losers. Once you select a door, the
game show host, who knows what is behind each door, does the following. First, whether or not
you selected the winning door, he opens one of the other two doors that he knows is a losing
door (selecting at random if both are losing doors). Then he asks you whether you would like to
switch doors. Which strategy should you use? Should you change doors or keep your original
selection, or does it not matter?

Solution: The probability you select the correct door (before the host opens a door and asks you
whether you want to change) is 1/3, because the three doors are equally likely to be the correct
door. The probability this is the correct door does not change once the game show host opens
one of the other doors, because he will always open a door that the prize is not behind.
The probability that you selected incorrectly is the probability the prize is behind one of the
two doors you did not select. Consequently, the probability that you selected incorrectly is 2/3.
If you selected incorrectly, when the game show host opens a door to show you that the prize is
not behind it, the prize is behind the other door. You will always win if your initial choice was
incorrect and you change doors. So, by changing doors, the probability you win is 2/3. In other
words, you should always change doors when given the chance to do so by the game show host.
This doubles the probability that you will win. (A more rigorous treatment of this puzzle can be
found in Exercise 15 of Section 7.3. For much more on this famous puzzle and its variations,
see [Ro09].) ▲