Transcript PowerPoint Presentation - Unit 1 Module 1 Sets, elements

Part 3 Module 5
The Multiplication Rule
Recall the following advice from our study of counting in
Part 1 Module 6:
Generally:
1. “or” means “add”
2. “and” means “multiply.”
We have seen, in Part 3 Modules 3 and 4, that the fact
#1 above carries over to probability:
P(E or F) = P(E) + P(F) for mutually exclusive events;
P(E or F) = P(E) + P(F) – P(E and F) in general.
We are about to see that fact #2 above also carries over
to probability.
“and” means “multiply”
In probability, just as in counting, “and” is
associated with multiplication.
We have the following fact that can be
used to calculate the probability of the
conjunction of two or more events:
For any events E, F
P(E and F) = P(E) × P(F, given E)
The Multiplication Rule
This fact is called the general
multiplication rule:
For any events E, F
P(E and F) = P(E) × P(F, given E)
In words: to find the probability that two events will both
occur, multiply the probability that the first event will
occur, times the probability that the second event will
occur, assuming that the first event did occur.
Example
At the Wee Folks Gathering there are
4 hobbits (H)
8 gnomes (G)
They will randomly select one person serve as Wolverine Groomer and a
different person to serve as Cat Washer.
1. What is the probability that the Wolverine Groomer is a hobbit (H1) and the
Cat Washer is a gnome (G2)?
A. .1667
B. .2424
C. .2222
D. None of these
2. What is the probability that both selectees are gnomes (G1 and G2)?
3. What is the probability that at least one selectee is a hobbit?
Solution
At the Wee Folks Gathering there are
4 hobbits (H)
8 gnomes (G)
They will randomly select one person serve as Wolverine Groomer and a
different person to serve as Cat Washer.
1. What is the probability that the Wolverine Groomer is a hobbit (H1) and the
Cat Washer is a gnome (G2)?
When they select the Wolverine Groomer, they have 12 people to choose
from, 4 of whom are hobbits, so P(H1) = 4/12. Assuming that the
Wolverine Groomer was a hobbit, when they select the Cat Washer they
have 11 people to choose from, 8 of whom are gnomes, so P(G2, given H1)
= 8/11. Thus, P(H1 and G2) = 4/12 × 8/11 = 32/132 = .2424
2. What is the probability that both selectees are gnomes (G1 and G2)?
P(G1 and G2) = 8/12 × 7/11 = 56/132 = .4242
Solution
At the Wee Folks Gathering there are
4 hobbits (H)
8 gnomes (G)
They will randomly select one person serve as Wolverine Groomer and a
different person to serve as Cat Washer.
3. What is the probability that at least one selectee is a hobbit?
The shortest way to answer this question is to use the complements rule,
applied to the answer to the previous question.
Since “at least one” is the opposite (complement) of “none,” the probability
that at least one selectee is a hobbit, is one minus the probability that
neither selectee is a hobbit, or
P(at least one selectee is a hobbit) = 1 – P(both selectees are gnomes)
= 1 – .4242 = .5758
Example
http://www.math.fsu.edu/~wooland/prob/prob22.html
History indicates that when Homerina the basketball player goes to the foul
line to shoot a pair of free throws, she will make the first shot 87% of the time.
If she makes the first shot, then she will make the second shot 78% of the
time. On the other hand, if she misses the first shot, then she will make the
second shot 46% of the time.
What is the probability that Homerina will miss the first shot and miss the
second shot?
A. 0.0598
B. 0.0939
C. 0.0909
D. 0.0702
E. 0.4698
Solution
History indicates that when Homerina the basketball player
goes to the foul line to shoot a pair of free throws, she will make the first shot 87% of the time.
If she makes the first shot, then she will make the second shot 78% of the time. On the other
hand, if she misses the first shot, then she will make the second shot 46% of the time. What
is the probability that Homerina will miss the first shot and miss the second shot?
P(miss both shots) = P(miss 1st shot) × P(miss 2nd shot, given missed 1st shot)
The data tells use that 87% of the time she makes the first shot; this means that she will miss
the first shot 13% of the time. P(miss 1st shot) =.13
That data tell us that if she misses the first shot, then 46% she will make the second shot, so,
if she misses the first shot, 54% of the time she will miss the second shot.
P(miss 2nd shot, given missed 1st shot) = .54
So, P(miss both shots) = .13 × .54 = .0702
Follow-up question
History indicates that when Homerina the basketball player
goes to the foul line to shoot a pair of free throws, she will make the first shot 87% of the time.
If she makes the first shot, then she will make the second shot 78% of the time. On the other
hand, if she misses the first shot, then she will make the second shot 46% of the time.
Use the answer to the previous question to answer this question:
What is the probability that she will make at least one of the two shots?
Answer:
Either (Make both shots) or (Make first and miss second) or (Miss first and make second)
.87 × .78
+
.87 × .22
+
.13 × .46
= .9892
Alternate solution
Since “At least one” is the opposite of “none,”
P(Make at least one shot) = 1 – P(Miss both shots)
= 1 – (.13 × .54)
= 1 – .0702 = .9892
Independent Events
Two or more events are independent if they don’t affect or influence one another.
More formally, E and F are independent events if the occurrence of E does not affect
the probability of F, and the occurrence of F does not affect the probability of E.
Example: suppose we have a red die and a green die. The experiment involves rolling each
die.
Let E be the event that when we roll the red die we get an even number, and let F be the
event that when we roll the green die we get a “2.”
These are independent events: the probability of one event is not affected by whether or
not the other event occurs. Suppose we want to find the probability that both events
occur…
The Multiplication Rule for Independent
Events
If E, F are independent events
P(E and F) = P(E) × P(F)
In words: to find the probability that two independent
events will both occur, multiply the probability that
the first event will occur, times the probability that
the second event will occur.
Referring to the previous question: The probability that
the red die roll will produce an even number is 3/6
and the probability that the green die roll will
produce a “2” is 1/6, so the probability that both
events occur is 3/6 × 1/6 = 3/36 = .0833
Pseudo-independent events
A survey a preschoolers indicates that 10% agree with the statement “broccoli
tastes good.” If two preschoolers are randomly selected, what is the
probability that both of them agree with the statement?
We must emphasize that there is not enough information to answer this question
exactly. In order to do so, we would have to know the exact size of the
population (N) and exactly how many agreed with the statement (A), in which
case
P(Both agree) = A/N × (A–1)/(N–1)
which is also the same as C(A,2)/C(N,2)
In a case like this, where an experiment involves selections from a large population of
unspecified size, we will approximate the answer by pretending that the
selections are independent, even though, strictly speaking, they aren’t.
Thus, P(both agree) = P(First person agrees) × P(Second person agrees)
= .1 × .1 = .01
Again, this answer (.01) is not the exact answer, but it is the best we can do working
only with the percent information, not knowing the size of the population. For
large populations, the resulting error is not significant.
Another example
Suppose a three-digit number is randomly created using digits from
this set:
{1, 3, 6, 8, 9}
What is the probability that the number has no repeated digits?
Solution: If we want to use the multiplication rule, we must understand
that what is important in this case is the following:
1.
Regardless of which digit is selected as the first digit, the second digit
must be different from the first digit. The probability that the second
digit is different from the first digit is 4/5.
2.
Regardless of which two digits were chose for the first and second
digits, the third digit must be different from both of the first two digits.
The probability that the third digit is different from both of the
first two digits is 3/5.
So, the probability that there are no repeated digits is 4/5 × 3/5 =
12/25= .48
This is the same as P(5,3)/(5×5×5×5), which is another correct way to
solve the problem.
“At least one” is the complement of “none”
New question: Suppose a three-digit number is randomly created
using digits from this set:
{1, 3, 6, 8, 9}
What is the probability that the number has at least one repeated
digit?
Solution: This is a much more complicated question to deal with directly,
but it is easy to answer if we realize that we can use the answer to the
previous question.
Since “at least one” is the opposite (complement) of “none”, we have
P(“at least one repeated digit”) = 1 – P(“no repeated digits”)
= 1 – .48 (.8 was the answer to the previous question)
= .52
Yet another example
A box contains a $1 bill, a $5 bill, a $10 bill, a $20 bill, a $50 bill and a $100
bill. Three bills are chosen without replace and their monetary sum is
determined. What is the probability that the monetary sum will be
$151.
Solution #1
An outcome in this experiment is a monetary sum obtained by adding the
values of the three bills, such as $16, $80, $35, or $151. Note that the
monetary is determined by which three bills are selected.
The number of equally-likely outcomes in this experiment is the number of
ways to choose three bills from a set of six bills: C(6,3) = 20.
There are twenty different, equally-likely, monetary sums.
Of the twenty different monetary sums, one of them “$151.”
So, P(“monetary sum = $151”) = 1/20 = .05
Alternative solution
If you would like to solve the previous problem by using the multiplication
rule instead of by solving a counting problem, here is the rationale:
The only way to get a monetary sum of $151 is if we select the $1 bill and
the $50 bill and the $100 bill.
When we select the “first” of the three bills, the probability that it will be the
$1 or the $50 or the $100 is 3/6.
Assuming that one of those three is selected, when we select the second
bill, the probability that it will one of the two “needed” bills that remain
will be 2/5.
Finally, assuming that the first two bills selected were two of our three
“needed” bills, the probability that the third “needed” bill will be
selected on the third draw is 1/4.
3/6 × 2/5 × 1/4 = 6/120 = .05