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:

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

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.

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 (_____) =

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

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

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.

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

Now consider just the numerators of the probabilities listed in the previous chart. P (_____) = 0 H 1 0 toss 1 toss P (___) = 0 H 1 1 P (___) = 1 H 2 tosses 1 2 1 P (___) = 0 H P (____) = 1 H P (____) = 2 H 1 3 3 1 3 tosses P (___) = 0 H P (___) = 1 H P (___) = 2 H P (___) = 3 H 4 tosses 1 4 6 4 1 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!

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.

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

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

Practice: Answer the following: 21 + 7 + 1 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

Practice: Answer the following: 1 + 6 + 15 + 20 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

Practice: Answer the following: 8 + 56 + 56 + 8 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

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

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

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

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

Finish the assignment: Worksheet & IC 118 – 122

Old Slides http://ptri1.tripod.com/ http://www.mathsisfun.com/pascals-triangle.html

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

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