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Pencil, highlighter, GP notebook,
textbook, assignment
Complete the first page of
“Allowance Dilemma #2” Worksheet.
total:
16
For your allowance each week, you have 2 choices:
A) Take $15
or…
B) Draw 2 bills, one at a time,
replacing it each time, from a bag
with three ones and two twenties.
What is your initial choice? (Circle one.) A
$ 1 $ 20
$ 20
$1 $1
or
B
Complete the area model below. Determine the sample space,
S, and the probability of each possible outcome.
1ST BILL +1 labels
+1
Sample Space:
$2, $21, $40
S = {_____________}
$1 $1 $1 $20 $20
2ND BILL
$1
2
2
2
21 21
$1
2
2
2
21 21
$1
2
2
2
21 21
$20 21 21 21 40 40
$20 21 21 21 40 40
+1 labels
+2 outcomes
Fraction
9
$2
P (____) =
25
12
$21
P (____) =
25
$40 =
P (____)
4
25
Decimal
= 0.36 +2
= 0.48 +2
= 0.16 +2
2. Determine the expected value (i.e., the long run average).
P($2)
P($21)
P($40)
0.36 $2 + (_____)(____)
0.48 $21 + (_____)(____)
0.16
$40
E(game) = (_____)(___)
+3
$17.20 +1
= _______
+1
B
Which choice, A or B, is a better deal in the long run? ______
Let’s consider another scenario…
total:
16
If you did not put the first bill back before selecting the second
bill, do you think the expected value would:
A) go up
b) go down
c) not change
Now the game has changed… we do not put the first bill back
before we draw the 2nd bill. Which option will you choose now?
A) Take $15
or…
B) Take 2 bills
$ 1 $ 20
$ 20
$1 $1
TREE DIAGRAM – Make a tree diagram for the new situation.
Label the probabilities on all branches.
1st Bill:
3
2
5 P ($20) = ___
5
P ($1) = ___
3
1st Bill:
5
2nd Bill:
If the first bill was a $1, then
$1
2
2
2 2nd Bill: 2
4
4
4 P ($20) = ___
4
P ($1) = ___
2nd Bill:
If the first bill was a $20, then
3
1
4 P ($20) = ___
4
P ($1) = ___
$1
$2
$20
$21
2
5
$20
3
4
1
4
$1
$21
$20
$40
Write the dollar totals at the end of each path. Multiply to find
the probability of each path.
Multiply to find the probability of each path.
 3  2  6
2 =   
P ($__)
 5   4  20
 3  2   2  3 
21 =       
P ($___)
 5  4   5  4 

6
6

20 20

12
20
 2  1   2
40 =    
P ($___)
 5   4  20
3
1st Bill: 5
2nd Bill:
2
4
$1
$2
$1
2
5
$20
2
4
3
4
1
4
$20
$21
$1
$21
$20
$40
DEPENDENCE Does the 1st bill have any effect on the
YES
outcome of the 2nd bill? ____
affects the 2nd bill, they are called __________
dependent
Since the 1st bill _______
events, that is, the outcome of one event affects the outcome of
a second event.
dependent events,
If A and B are __________
P(B following A) That is, we
P(A)  _______________.
then P(A and B) = _____
multiply the probabilities.
EXPECTED VALUE – Compute the expected value.
P($2)
P($21)
P($40)
0.3 $2 + (____)(____)
0.6 $21 + (____)(____)
0.1 $40
E(game) = (____)(___)
$17.20
= _______
B
Which choice, A or B, is a better deal in the long run? ______
Did the expected value change from the original problem?
YES
or
NO
PM – 40
Suppose you roll two dice, one red and one blue, and get a
sum of 10.
a) List the different ways this can occur.
red 5, blue 5
red 6, blue 4
red 4, blue 6
b) Sketch an area diagram and shade these possibilities.
1
2
3
4
5
6
c) What is the probability of
getting a sum of 10?
P (sum = 10) = 3
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10 11 There are 3 ways to get 10, one of
6
7
8
9
10 11 12
36
d) Suppose you know the sum is
10 but not what is on each side.
Explain why the probability that
you rolled two 5’s would be 13 .
them is two 5’s.
PM – 41
Suppose you roll two dice and the sum is more than 8.
a) Shade the squares on an area diagram where this outcome could
occur.
b) What is the probability that both dice show the same number?
P (same # given that the sum is more than 8) = 2
10
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10 11
6
7
8
9
10 11 12
c) What is the probability that
exactly one 6 is showing?
P (exactly one 6 given that
6
the sum is more than 8) = 10
d) What is the probability that
at least one 5 is showing?
P (at least one 5 given that
5
the sum is more than 8) = 10
PM – 42
A spinner comes up blue, red, and green with a probability
of 1 for each color.
3
a) Sketch an area diagram for spinning twice.
b) Shade the region on your area diagram that
corresponds to getting the same color twice.
Spin #1
B
R
G
B BB RBGB
Spin #2
R BR RRGR
G B G R GG G
c) What is the probability that
both spins give the same color?
P (same color) = 3
9
d) If you know that you got the
same color twice, what is the
probability the color was blue?
P (blue | same color twice) = 1
3
A spinner comes up red 25% of the time and green 25% of
the time. The rest of the time it lands on blue.
a) Draw an area diagram for spinning twice, and
shade the region on your area diagram corresponding
to getting the same color twice.
PM – 43
What are the dimensions of the whole diagram?
The dimensions are 1 x 1.
Spin #1
Spin #2
B
B
R
G
B BBBB RBGB
B BBBB RBGB
R B RB R R RG R
G B G B G R GG G
b) What is the probability that both
spins give the same color?
P (same color) = 6
16
c) If you know that you got the same
color twice, what is the probability
the color was blue?
P (blue | same color twice)  4
6
ALTERNATE AREA DIAGRAM FOR PM – 43
A spinner comes up red 25% of the time and green 25% of
the time. The rest of the time it lands on blue.
a) Draw an area diagram for spinning twice, and
shade the region on your area diagram corresponding
to getting the same color twice.
PM – 43
What are the dimensions of the whole diagram?
The dimensions are 1 x 1.
Spin #2
Spin #1
R
Blue
G
1
2
1
4
1
4
B 1
1
4
1
8
1
8
R 1
1
8
1
8
1
16
1
16
1
16
1
16
2
4
1
G 4
PM – 43
b) What is the probability that both spins give the same color?
Spin #2
Spin #1
R
Blue
G
1
2
1
4
1
4
B 1
1
4
1
8
1
8
R 1
1
8
1
8
1
16
1
16
1
16
1
16
2
4
1
G 4
P (same color) = 1  1  1
4 16 16
 4  1  1
16 16 16
 6
16
c) If you know that you got the same color twice, what is
the probability the color was blue?
PM – 43
Spin #1
R
Blue
Spin #2
1
2
1
4
G
1
4
B 1
1
4
1
8
1
8
R 1
1
8
1
8
1
16
1
16
1
16
1
16
2
4
1
G 4
P (blue | same color twice)
1  16
4
6
=
6
16
16  6
=
1  16
4 6
=
2
3
PM – 41
e) In part (d) you calculated the conditional probability that a five will
show given the sum on the dice is greater than 8. Here is another
way to think of the problem:
Event A is sums that contain at least one 5.
Event B is sums greater than 8.
Event A
6
7
8
9
6 7 8 9 10 11
11
Event B
9
9 10
9 10 11
9 10 11 12
PM – 41
Event A is sums that contain at least one 5.
Event B is sums greater than 8.
Event AB is the intersection of events A and B, that is, sums
that contain at least one 5 and are greater than 8.
6
7
8
6 7 8
Event AB
9
9
10
9 10 11 9 10 12
11
Then we can compute P(A | B), the
conditional probability that at
least one 5 showing given that the
sum was greater than 8, by using
the probability P(AB) and P(B).
5
P(A and B)
P(A | B) =
= 36
P(B)
10
36
“given that”
5 36
5

=

36 10 10