pencil red pen highlighter GP notebook calculator Complete

pencil, red pen, highlighter, GP notebook, calculator 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:

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

Pascal’s Triangle. This triangle is called ____ 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 1 If you toss a coin ___time, there are 2 __ possibilities for the 0 or ___, 1 both ____ equally likely. number of heads, ___ P (_____) 0 H = P (_____) 1 H = 2 If you toss a coin _____ times, write out the sample space. HH, HT, TH, TT S = {________} 3 possibilities for the number of heads, but ____ not all There are ___ equally likely. _______ P (_____) 0 H = P (_____) 1 H = 2 H P (_____) =

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

4 times, there are _____ 24 = _____ 16 different If you toss a coin ___ outcomes in the sample space. S= 5 possibilities for the number of heads, ranging There are ___ 0 to ____ 4 heads. from ____ 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 ____ 0 times, there is ____ 1 possibility for the number of heads, only ____ 0 heads. 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: 0 0 1 Number of tosses 2 3 4 P(# of heads) 1 2 3 4

ÞNow consider just the numerators of the probabilities listed in the previous chart. 0 toss 1 toss P (_____) 0 H =1 P (___) 0 H = 1 P (___) 1 H = 1 2 tosses P (___) 0 H = 1 P (____) 1 H = 2 P (____) 2 H = 1 3 tosses P (___) 0 H = 1 P (___) 1 H = 3 P (___) 2 H = 3 P (___) 3 H = 1 tosses P 4(___) 0 H = 1 P (___) 1 H = 4 P (___) 2 H = 6 P (___) 3 H = 4 P (___) 4 H = 1 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 5 32 in the sample space? ___ = 2___ Compute the probabilities if a coin is tossed five times: P (0 H) = P (1 H) = P (2 H) = P (3 H) = P (4 H) = P (5 H) = 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 0 tosses 1 1 toss 1 2 tosses 1 3 tosses 1 4 tosses 1 5 tosses 1 6 tosses 7 tosses 8 tosses 1 Sum 1 1 1 8 3 5 7 6 15 56 1 4 10 20 35 8 1 5 15 35 70 2 4 1 3 10 21 28 2 4 6 1 1 6 21 56 16 32 1 7 28 64 1 8 128 1 256

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

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

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

Practice: Answer the following: 6) P (7 H in 15 tosses) = 0 tosses Uh oh… the triangle stops at 8 tosses. 1 What do we do? ? ? 1 toss 1 2 tosses 1 1 2 Extending Pascal’s triangle to 3 tosses 1 3 3 15 tosses is not practical. 4 tosses 1 4 6 at a shortcut. 5 tosses Let’s look 1 5 10 10 6 tosses 7 tosses 8 tosses 1 Sum 1 1 1 6 7 8 15 21 28 20 35 56 1 1 4 8 1 5 15 35 70 2 4 16 1 6 32 1 64 What are we 21 to do? ? ? 7 1 going 56 28 8 128 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 2) P (5 H in 8 tosses) = 56 256 3) P (at least 5 H in 7 tosses) = 7 C 5 + 7 C 6 + 7 C 7 (5 or more heads) 27

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

P (r H in n tosses) = 7) P (3 H in 10 tosses) = 120 1024 8) P (4 H in 16 tosses) = 1820 65536 9) P (at least 4 H in 6 tosses) = 10) P (6 H in 11 tosses) = 11) P (8 H in 12 tosses) = 6 C 4 + 6 C 5 + 6 C 6 26 = 22 64 462 2048 495 4096 12) P (at most 2 H in 9 tosses) = 9 C 0 + 9 C 1 + 9 C 2 29 = 46 512

Finish the assignment: Worksheet & IC 118 – 122

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

P (r H in n tosses) = 3) P (at least 5 H in 7 tosses) = 7 C 5 + 7 C 6 + 7 C 7 (5 or more heads) 27 4) P (at most 3 H in 6 tosses) = (0 to 3 heads) 5) P (odd # of H in 8 tosses) = 6) P (7 H in 15 tosses) = 15 C 7 215 6 C 0 + 6 C 1 + 6 C 2 + 6 C 3 26 8 C 1 + 8 C 3 + 8 C 5 + 8 C 7 28 6435 = 32768