1. pencil, red pen, highlighter, GP notebook, calculator
U11D9
Have out:
Bellwork:
Complete the following on today’s worksheet:
Use your calculator to compute the values for each combination. Write the answer in the box, and look for patterns.
Complete the last 2 rows using any patterns you have noticed.
Write the sum for each row in the column on the right.
total:

2. Bellwork:
Sum 1 1
+1 for each correct row 1 1 2
+1 for each correct sum 1 2 1 4
total: 1 3 3 1 8 1 4 6 4 1 16 1 5 10 10 5 1 32 1 6 15 20 15 6 1 64 1 7 21 35 35 21 7 1
128 1 8 28 56 70 56 28 8 1
256
Sum of the nth row is ____ 2n

3. This triangle is called ________ Triangle.
Pascal’s
List some patterns:
 The triangle can be built using combinations.
 The sums are powers of 2.
 Each row starts and ends with 1.
 The triangle is symmetrical.
 Excluding the 1’s, each number in a row is the sum of the two numbers above it.

4. Probabilities of Tossing Coins
If you toss a coin ___time, there are __ possibilities for the number of heads, ___ or ___, both ________ likely. 1 2 1
equally
P (_____) =
0 H
P (_____) =
1 H
If you toss a coin _____ times, write out the sample space.
S = {________________} 2
HH,
HT,
TH, TT
There are ___ possibilities for the number of heads, but ____ all _______ likely. 3
not
equally
2 H
P (_____) =
0 H
P (_____) =
1 H
P (_____) =

5. If you toss a coin ___ times, write out the sample space.3
HHH,
HHT,
HTH,
HTT,
THH,
THT,
TTH,
TTT
There are ___ possibilities for the number of heads, but ____ all _______ likely 4
not
equally
2 H
P (_____) =
0 H
P (_____) =
1 H
P (_____) =
P (_____) =
3 H

6. If you toss a coin ___ times, there are _____ = _____ different outcomes in the sample space.4 24 16
S =
There are ___ possibilities for the number of heads, ranging
from ____ to ____ heads. 5 4
Try these on your own:
P (___) =
0 H
P (___) =
1 H
P (___) =
2 H
P (___) =
3 H
P (___) =
4 H

7. If you toss a coin ____ times, there is ____ possibility for the number of heads, only ____ heads.1
P (_____) =
0 H 1
We could continue to list the probabilities of tossing coins 5 times, 6 times, 7 times, etc., but this gets very tedious.
Let’s see if there are some patterns emerging.

8.  Let’s put it all together:
P(# of heads) 1 2 3 4
Number of tosses

9. Now consider just the numerators of the probabilities listed in
the previous chart.
P (_____) =
0 H 1
0 toss
1 toss
P (___) =
0 H
P (___) =
1 H
2 tosses
P (___) =
0 H
P (____) =
1 H
P (____) =
2 H
3 tosses
P (___) =
0 H
P (___) =
1 H
P (___) =
2 H
P (___) =
3 H
4 tosses
P (___) =
0 H
P (___) =
1 H
P (___) =
2 H
P (___) =
3 H
P (___) =
4 H
The numerators are Pascal’s Triangle!
The denominators are the sums of each row!

10. Based on the patterns we discussed, answer the following:
If you toss a coin 5 times, how many different outcomes are there in the sample space? ___ = ___ 25 32
Compute the probabilities if a coin is tossed five times:
P (0H) =
P (1H) =
P (2H) =
P (3H) =
P (4H) =
P (5H) =
Complete Practice #1 – 6. Use Pascal’s Triangle to answer most of these.

11. Practice: Answer the following:15
1) P (4 H in 6 tosses) = 64
Sum
0 tosses 1 1
1 toss 1 1 2
2 tosses 1 2 1 4
3 tosses 1 3 3 1 8
4 tosses 1 4 6 4 1 16
5 tosses 1 5 10 10 5 1 32
6 tosses 1 6 15 20 15 6 1 64
7 tosses 1 7 21 35 35 21 7 1
128
8 tosses 1 8 28 56 70 56 28 8 1
256

12. Practice: Answer the following:56
2) P (5 H in 8 tosses) =
256
Sum
0 tosses 1 1
1 toss 1 1 2
2 tosses 1 2 1 4
3 tosses 1 3 3 1 8
4 tosses 1 4 6 4 1 16
5 tosses 1 5 10 10 5 1 32
6 tosses 1 6 15 20 15 6 1 64
7 tosses 1 7 21 35 35 21 7 1
128
8 tosses 1 8 28 56 70 56 28 8 1
256

13. Practice: Answer the following:
3) P (at least 5 H in 7 tosses) =
128
(means 5 or more heads)
Sum
0 tosses 1 1
1 toss 1 1 2
2 tosses 1 2 1 4
3 tosses 1 3 3 1 8
4 tosses 1 4 6 4 1 16
5 tosses 1 5 10 10 5 1 32
6 tosses 1 6 15 20 15 6 1 64
7 tosses 1 7 21 35 35 21 7 1
128
8 tosses 1 8 28 56 70 56 28 8 1
256

14. Practice: Answer the following:
4) P (at most 3 H in 6 tosses) = 64
(means 0 to 3 heads)
Sum
0 tosses 1 1
1 toss 1 1 2
2 tosses 1 2 1 4
3 tosses 1 3 3 1 8
4 tosses 1 4 6 4 1 16
5 tosses 1 5 10 10 5 1 32
6 tosses 1 6 15 20 15 6 1 64
7 tosses 1 7 21 35 35 21 7 1
128
8 tosses 1 8 28 56 70 56 28 8 1
256

15. Practice: Answer the following:
5) P (odd # of H in 8 tosses) =
256
(means 1, 3, 5, or 7 heads)
Sum
0 tosses 1 1
1 toss 1 1 2
2 tosses 1 2 1 4
3 tosses 1 3 3 1 8
4 tosses 1 4 6 4 1 16
5 tosses 1 5 10 10 5 1 32
6 tosses 1 6 15 20 15 6 1 64
7 tosses 1 7 21 35 35 21 7 1
128
8 tosses 1 8 28 56 70 56 28 8 1
256

16. Uh oh… the triangle stops at 8 tosses.
Practice: Answer the following:
6) P (7 H in 15 tosses) =
Uh oh… the triangle stops at 8 tosses.
Sum
0 tosses 1
What do we do???
What are we going to do??? 1
1 toss 1 1 2
2 tosses 1 2 1 4
Extending Pascal’s triangle to 15 tosses is not practical.
3 tosses 1 3 3 1 8
4 tosses 1 4 6 4 1 16
Let’s look at a shortcut.
5 tosses 1 5 10 10 5 1 32
6 tosses 1 6 15 20 15 6 1 64
7 tosses 1 7 21 35 35 21 7 1
128
8 tosses 1 8 28 56 70 56 28 8 1
256

17. Recall from today’s lesson that Pascal’s Triangle can be formed from combinations.
Each of the practice problems could have been solved using the pattern:
P (r H in n tosses) =
Let’s go back and verify that the pattern works for #1 – 3 using your calculators. 15
1) P (4 H in 6 tosses) = 64 56
2) P (5 H in 8 tosses) =
256
7C5
+ 7C6
+ 7C7
3) P (at least 5 H in 7 tosses) =
(5 or more heads) 27

18. Finish the assignment:
P (r H in n tosses) =
Let’s answer #6:
15C7
6435
6) P (7 H in 15 tosses) = =
215
32768
Finish the assignment:
Worksheet & IC 118 – 122

19. P (r H in n tosses) =
120
7) P (3 H in 10 tosses) =
1024
1820
8) P (4 H in 16 tosses) =
65536
6C4
+ 6C5
+ 6C6 22
9) P (at least 4 H in 6 tosses) = = 26 64
462
10) P (6 H in 11 tosses) =
2048
495
11) P (8 H in 12 tosses) =
4096
9C0
+ 9C1
+ 9C2
12) P (at most 2 H in 9 tosses) = 46 = 29
512

20. Finish the assignment:
Worksheet & IC 118 – 122

21. Old Slides http://ptri1.tripod.com/

22. Pascal’s Triangle
Sum
0 toss 1
1 toss
2 tosses
3 tosses
4 tosses
5 tosses
6 tosses
7 tosses
8 tosses 1 2 1 2 1 4 1 3 3 1 8 1 4 6 4 1 16 1 5 10 10 5 1 32 1 6 15 20 15 6 1 64 1 7 21 35 35 21 7 1
128 1 8 28 56 70 56 28 8 1
256
Sum of the nth row is ____ 2n

23. P (r H in n tosses) =
7C5
+ 7C6
+ 7C7
3) P (at least 5 H in 7 tosses) =
(5 or more heads) 27
6C0
+ 6C1
+ 6C2
+ 6C3
4) P (at most 3 H in 6 tosses) =
(0 to 3 heads) 26
8C1
+ 8C3
+ 8C5
+ 8C7
5) P (odd # of H in 8 tosses) = 28
15C7
6435
6) P (7 H in 15 tosses) = =
215
32768