Download Example 3, Pg. 253, #7

Chapter 14. More about Chance
In Chapter 13 we learned how to find probabilities of events A and B both occurring,
using the multiplication rule. In this chapter again we find probabilities, though this time
of events A or B occurring, or in other words, that at least one occurs.
Key Terms
Mutually Exclusive – Two things are mutually exclusive when the occurrence of one
prevents the occurrence of the other: one excludes the other.
The Addition Rule – Finds the chance that at least one of two things happens. More
specifically, if two events are mutually exclusive, we add the chances.
Key Formulas
Do an experiment. The set of all possible outcomes is O. An EVENT is any subset A of O. An even is a
collection of possible outcomes.
1. Assuming that each outcome is equally likely, the probability of A, written P(A) is
P( A) 
2.
3.
4.
5.
6.
7.
number _ of _ outcomes _ in _ A
number _ of _ outcomes _ in _ O
We always have 0 <=P(A)<=1. Probabilities are always between zero and one.
The probability of the opposite event, not A, written A^c, is P(A^c)=1-P(A).
The probability of “either A occurs OR B occurs, OR BOTH occur” is
P( A _ or _ B )  P( A)  P( B )  P( A _ AND _ B ) .
The SUM RULE: A and B are MUTUALLY EXCLUSIVE if they cannot occur together. One can
happen, or the other can happen, but they can’t both happen at the same time. When they are
mutually exclusive P(A_or_B)=P(A)+P(B).
The MULTIPLICATION RULE: A and B are INDEPENDENT if P(A AND B)=P(A) X P(B).
The probability of A given that B has occurred is written P(A|B).
P(A|B)=P(A AND B)/P(B). A and B are independent if and only if P(A|B)=P(A).
LISTING THE WAYS
Some times the easiest way to find the chances is by listing all the possible outcomes and
using the formula P(A)= number of ways A can occur/ all possible outcomes.
EXAMPLE: #9 p 253 One ticket is drawn from box A: 1 2 3 and one from B: 1 2 3 4.
a) The chance that the number from A is larger than the one from B will be written
P(A>B). Now this is easy if you write all the possibilities:
O={1-1,1-2,1-3,1-4,2-1,2-2,2-3,2-4,3-1,3-2,3-3,3-4}
The fist number is the number drawn from A and the second from B. Now the event
A>B = {2-1,3-1,3-2} so P(A>B)=3/12.
b) P(A=B)=3/12.
c) P(A<B)= P(1-2,1-3,1-4,2-3,2-4,3-4)=6/12.
By Listing the ways in the experiment of trowing a PAIR of dice, you can see that the
chances
P(roll a 7)=6/36, P(roll a 6)=P(1-5,2-4,3-3,5-1,4-2)=5/36.
Lisitng the ways works only if the number of possible outcomes is not too big.
MUTUALLY EXCLUSIVE AND THE ADDITION PRINCIPLE
In the above definition for the addition rule, I claim that the chances can be added
if two events are mutually exclusive. What if two events are not mutually exclusive? In
this case we cannot add the chances because our sum will be too big. Let’s see why,
through an example.
Example
Find the probability of drawing a playing card that is an ace OR a diamond.
Solution
First consider whether the two events are mutually exclusive. By definition, one must
exclude the other. Ask yourself: if the first card is an ace does that mean it can’t be a
diamond? If it does then the two events are mutually exclusive. Since a deck of cards
contains an ace that is a diamond, our answer to the question is “no”. Therefore, these
events are not mutually exclusive and you cannot directly apply the definition above.
Write A={card is an Ace} and B={card is a diamond}. You want P(A OR B). A and B
are NOT mutually exclusive. If you ignored this and used the sum rule you would get
P(A OR B) = P(A) + P(B) = 4/52 + 13/52 = 17/52. This is WRONG. You double counted
the one card which is an Ace AND a diamond. So you should use the rule when they
aren’t mutually exclusive: P(A OR B)=P(A)+P(B)-P(A AND B)=4/52+13/521/52=16/52.
P(A and B)=There is only one ace of diamonds in the deck and so this probability would
be 1/52.
The solution is then: 4/52 + 13/52 – 1/52 = 16/52 = 4/13
Probability of (A or B)
Addition Rule
Are (A and B)
mutually exclusive?
No
Probability of (A or B) equals the
Probability of A plus the Probability of B
Minus the Probability of (A and B)
Yes
Probability of (A or B) equals the
Probability of A plus Probability of B
Example
A box contains 10 red balls numbered one through ten and 10 black balls numbered one
through ten. What is the chance of picking the red ball numbered 3 or the black ball
numbered 3?
Solution
If the ball picked is red, then it can’t be black and so these two events are mutually
exclusive. Therefore, applying the addition rule for two mutually exclusive events, we
add the chances that each happen. The probability of picking the red ball numbered 3 is
1
1
20 . The probability of picking the black ball numbered 3 is 20 . The chance of picking
either the red ball numbered 3 or the black ball numbered three is additive:
1
1
2
1



20 20 20 10 .
MUTUALLY EXCLUSIVE VS. INDEPENDENCE
Sometimes students mistake mutually exclusiveness and independence as the
same thing. Both describe the relationship of a pair, but notice how these relationships
are different:
(b) Two things are mutually exclusive if the occurrence of one prevents the
occurrence of the other from happening.
(c) Two events are independent if the occurrence of one does not change the
chances for the other.
SHALL I ADD OR MULTIPLY?
Students also often get confused in deciding when to multiply and add when
figuring probabilities. One must realize that the two rules solve different problems.
Addition:
(a) The addition rule finds the chance that at least one of two things happens.
(b) Mathematically speaking, the addition rule finds P(A or B).
(c) Adding the probabilities of two events requires them to be mutually exclusive.
Multiplication:
(a) The multiplication rule finds the chance that two things both happen.
(b) Mathematically speaking, the multiplication rule finds P(A and B)
(c) Multiplying the unconditional probabilities of two events requires them to be
independent. For dependent events, the multiplication rule uses conditional
probabilities
Now here are some mixed examples of both principles.
Example 3, Pg. 253, #7
Solution
This problem can be approached in a manner similar to the way that the Paradox of
Chevalier de Méré was solved.
When it is difficult to calculate a probability directly, a good technique is to subtract its
opposite from the number 1.
Probability of Event A = 1 – (Probability of the Opposite of Event A)
In this exercise, the probability of drawing a 2 is two out of five, whereas the probability
of not drawing a 2 is three out of five. We cannot add the probabilities because the
events are not mutually exclusive. Independence can be applied because the numbers are
drawn with replacement. Since a number is drawn four times, then the answer is
Probability of drawing a 2 at least once = 1 – Probability of not drawing a 2
 3
Probability of drawing a 2 at least once = 1 -  
5
4
Example, Pg. 253, #11
Solution
(a) The probability that the first card drawn is a diamond =
that the second card drawn is a diamond =
13
. The probability
52
12
. And the probability that the third
51
11
. Since each drawing is independent of the other,
50
we can use the multiplication rule. Therefore:
card drawn is a diamond =
13 12 11 1716
  =
 1%
52 51 50 132600
(b) There are 52 cards and 13 of them are diamonds and 39 of cards are not. This
means that the chance of not picking a diamond is 39 out of 52. By a similar
calculation to part (a), use the multiplication rule:
39 38 37 54834
 

 41%
52 51 50 132600
(c) From part (a), the chance of getting all diamonds is 1%. Consequently, the
chance of not getting all diamonds is 99%.