Download Conditional Probability and Independent Events Monday, July 20

Conditional Probability and Independent Events
Monday, July 20
Often, counting problems have extra conditions that modify the relative sample space. For example, if
two cards are drawn without replacement from a deck, what is the probability that the second card drawn
is an ace, given the first card drawn was not an ace. Our new sample space will be every event that
follows the rule: the first card drawn was not an ace. Then we are looking for how many times in our new
sample space the second card drawn is an ace. These problems are known as conditional probability.
We compute the probability that an event B happens, given that the event A has already occurred. This
probability, denoted P (B|A), is called the conditional probability of B given A.
of elements in E
Recall that P (E) = number
. In a conditional probability, our sample space is reduced, and the
number of elements in S
number of times E can occur is limited by the outcomes in the sample space. So
P (B|A) =
n(A ∩ B)
number of elements in A ∩ B
=
number of elements in A
n(A)
By dividing numerator and denominator by n(S), the number of elements in S, we have
P (B|A) =
n(A∩B)
n(S)
n(A)
n(S)
=
P (A ∩ B)
P (A)
Example 1: A pair of fair dice is rolled. What is the probability that the sum of the numbers falling
uppermost is 7 if it is known that one of the numbers is a 5?
Let A denote the event that the sum of the numbers is 7, and let B denote the event that one of the
numbers is a 5. Then
A = {(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)}
B = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)}
A ∩ B = {(5, 2), (2, 5)}
Since the dice are fair, each outcome is equally fair, so
P (A ∩ B) =
2
11
, P (B) = .
36
36
Thus the probability that we are looking for is
P (A|B) =
1
2
36
11
36
=
2
.
11
Exercise 1: In a test conducted by the U.S. Army, it was found that 1000 new recruits (600 men and 400
women), 50 of the men and 4 of the women were red-green color-blind. Given that a recruit is selected at
random from this group is red-green color-blind, what is the probability that the recruit is male?
Sometimes, we know the probability of B occurring, given A has already occurred, and we wish to know
the probability of A and B occurring. Then we can rewrite out equation for P (B|A) so that we have
P (A ∩ B) = P (A) · P (B|A).
Sometimes it is helpful to use tree diagrams to help describe a problem. Consider the following example.
Example 2: A box contains eight 9-volt batteries, of which two are known to be defective. The abtteries
are selected one at a time without replacement and tested until a nondefective one is found. What is the
probability that the number of batteries tested is (a) One? (b) Two? (c) Three?
Solution:
With the aid of the tree diagram, we see that (a) is
6
8
= 34 , (b) is ( 28 )( 67 ) =
3
,
14
and (c) is ( 28 )( 17 )(1) =
1
.
28
Two events are called independent if the outcome of one does not depend on the other. For example,
when rolling a die twice, the second roll is not dependent on the first roll. However, when drawing two cards
out of a deck without replacement, the first card drawn changes the size of the sample space for the second
card. Precisely, two events A and B are called independent if P (A|B) = P (A) and P (B|A) = P (B).
Also, two events are independent if and only if P (A ∩ B) = P (A)P (B). Note that two mutually exclusive
events are not independent, since the first event occurring means the second event cannot occur at all!
Exercise 2: Determine whether E and F are (a) mutually exclusive and (b) independent if:
a. P (E) = .5, P (F ) = .4, and P (E ∩ F ) = .2.
b. P (E) = .5, P (F ) = .7, and P (E ∩ F ) = .3.
c. P (E) = .1, P (F ) = .3, and P (E ∩ F ) = 0.
2