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Key Stone Problem… Key Stone Problem… next Set 6 © 2007 Herbert I. Gross

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You will soon be assigned five problems to test whether you have internalized the material in Lesson 6 of our algebra course. The Keystone Illustration below is a prototype of the problems you'll be doing. Work out the problem on your own. Afterwards, study the detailed solutions we've provided. In particular, notice that several different ways are presented that could be used to solve the problem. Instructions for the Keystone Problem next © 2007 Herbert I. Gross

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As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. next © 2007 Herbert I. Gross

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next (a) A penny, a nickel, a dime and a quarter are flipped simultaneously. In how many ways can the coins turn up “heads” or “tails”? Keystone Illustration for Lesson 6 Answer: There are 16 ways. © 2007 Herbert I. Gross next

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Answer: 16 Solution for Part a: Each coin can turn up in either of two ways, either “heads” or “tails”. Each time we add a coin to the collection, we double the number of outcomes. Namely, each of the previous outcomes is obtained twice, once if the additional coin turns up heads and once if the additional coin turns up tails. next © 2007 Herbert I. Gross

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Solution for Part a: If the penny is flipped by itself, there are 2 possible outcomes. In exponent form 2 1 = 2. If the penny and nickel are flipped by themselves there are 2 × 2 or 4 outcomes. In exponent form 2 2 = 4. next © 2007 Herbert I. Gross

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Solution for Part a: If the penny, nickel, and dime are flipped by themselves, there are 2 × 4 or 8 outcomes. In exponent form 2 3 = 8 If the penny, nickel, dime, and quarter are flipped by themselves there are 2 × 8 or 16 outcomes. In exponent form 2 4 = 16. next © 2007 Herbert I. Gross

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next (b) How does the answer to part (a) change if a penny, nickel, dime, quarter, and half dollar are flipped?. Keystone Illustration for Lesson 5 Answer: The number of ways doubles. Therefore, there are 32 ways. © 2007 Herbert I. Gross next

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Solution for Part b: In part (a) we pointed out that every time one more coin is added to the collection of coins that are being flipped the number of is twice the original amount. We already know from part (a) that there are 16 outcomes when the four coins were flipped. Hence when the 5th coin is added, there will be 2 × 16 or 32 outcomes. In exponent form 2 5 = 32. next © 2007 Herbert I. Gross

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next (c) Twenty pennies are flipped simultaneously. In how many ways can the coins turn up “heads" or “tails”? Keystone Illustration for Lesson 5 Answer: There are 2 20 or 1,048,576 ways. © 2007 Herbert I. Gross next

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Solution for Part c: Based on what we saw in parts (a) and (b) of this problem it should becoming clear that if there are n coins being flipped there are 2 n possible outcomes. next © 2007 Herbert I. Gross

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Solution for Part c: For example… next © 2007 Herbert I. Gross Number of coins (n)Number of outcomesExponential notation 122121 242 382323 4162424 5322525 6642626 71282727

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Solution for Part c: So with 20 coins the number of possible outcomes is 2 20, which is a far less cumbersome way for expressing… next © 2007 Herbert I. Gross 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 next

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© 2007 Herbert I. Gross Note In other words, if 20 “fair” coins are flipped, there is 1 chance in 1,048,576 that all 20 coins will turn up heads. So if 20 heads do turn up, either the coins are not all “fair”, or else you are seeing an event for which the odds are 1,048,575 to 1 against. next A similar situation exists with respect to a true false test. Namely if there are 20 questions and you rely solely on guessing, the odds against you getting all 20 answers correct are 1,048,575 to 1.

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next © 2007 Herbert I. Gross Note With respect to part (c) we could multiply 2 by itself the required number of times and finally arrive at 1,048,576 ways. Namely… next n2n2n n2n2n 12112,048 24124,096 38138,192 4161416,384 5321532,768 6641665,536 712817131,072 825618262,134 951219524,288 101,024201,048,576

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next © 2007 Herbert I. Gross Mathematically, an answer such as 2 20 is absolutely correct. However most people are more comfortable seeing the answer expressed in the more traditional place value system. This is where the use of the calculator is quite helpful. Namely if the calculator has an x y key we simply use the following sequence of key strokes… next 2xyxy 20= and the answer 1,048,576 appears in the display window. 1,048,576

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next © 2007 Herbert I. Gross While it is tedious to multiply twenty factors of 2, there are a few shortcuts, thanks to such rules as the associative and commutative properties of multiplication. For example, suppose we have already computed that 2 5 = 32. If we look at 2 10 in its expanded form; that is, as… 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 next by the associative property we may rewrite it as… (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)

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© 2007 Herbert I. Gross We know from our chart that 2 5 = 32. Hence we may rewrite … next ( 2 × 2 × 2 × 2 × 2 ) × ( 2 × 2 × 2 × 2 × 2 ) 4 4 8 8 16 32 = 1,024 as And if the product of ten 2's is 1,024 the product of twenty 2's is… 1,024 × 1,024 = 1,048,576

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© 2007 Herbert I. Gross We saw in our Lesson 6 presentation that to list the 8 outcomes when a dime, nickel, and a penny were flipped, it would require using 8 lines. By way of review … Dime Outcome #1 Outcome #2 Outcomes NickelPenny Outcome #3 Outcome #4 Outcome #5 Outcome #6 Outcome #7 Outcome #8 heads tails tail heads tails heads tails heads tails heads tails heads tails heads tails heads tails next

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© 2007 Herbert I. Gross The number of rows doubles each time we add another coin. So to list the 16 outcomes that occur if four coins are flipped, we would need 16 rows. That is, each of the above 8 outcomes could still occur regardless of whether the fourth coin turned up heads or tails. In terms of a chart… next

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© 2007 Herbert I. Gross dimenickelpennyquarter heads tails heads tailsheads tails headstailsheads Outcome 1 Outcome 2 Outcome 3 Outcome 4 Outcome 5 Outcome 6 Outcome 7 Outcome 8 Outcome 9 Outcome 10 Outcome 11 Outcome 12 Outcome 13 Outcome 14 Outcome 15 Outcome 16 headstailsheadstails headstails heads tails heads tailsheads tails headstailsheads tailsheadstails heads tails headstails heads tails next

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© 2007 Herbert I. Gross While making a list of the outcomes can quickly become tedious as the number of coins increases, a lot of information can be gleaned from the explicit list. For example, there is a tendency for people to believe that if there are 4 coins there's a 50-50 chance that on a given flip there will be 2 heads and 2 tails. next

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© 2007 Herbert I. Gross However, if we look at the list of possible outcomes, we see that there are only 6 of the 16 outcomes where this happens; namely for Outcome #'s 4, 6, 7, 10, 11 and 13. What is true is that of all the possible outcomes 2 heads and 2 tails is more likely than any other outcome.

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next © 2007 Herbert I. Gross However obtaining a distribution of 3 of one kind and 1 of the other is more likely than obtaining the same number of heads as tails. More specifically, there are 4 outcomes that yield 3 heads and 1 tail and another 4 outcomes that yield 3 tails and 1 head. In terms of a chart… DistributionNumber of Ways 4 heads, 0 tails1 3 heads, 1 tails4 2 heads, 2 tails6 1 heads, 3 tails4 0 heads, 1 tails1 Total Outcomes 16

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© 2007 Herbert I. Gross Hence 3 of one kind and 1 of the other will appear 8 times… dimenickelpennyquarter heads tails heads tailsheads tails headstailsheads Outcome 1 Outcome 2 Outcome 3 Outcome 4 Outcome 5 Outcome 6 Outcome 7 Outcome 8 Outcome 9 Outcome 10 Outcome 11 Outcome 12 Outcome 13 Outcome 14 Outcome 15 Outcome 16 headstailsheadstails headstails heads tails heads tailsheads tails headstailsheads tailsheadstails heads tails headstails heads tails and 2 of one kind and 2 of the other will appear only 6 times. next

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© 2007 Herbert I. Gross Notice that constructing the chart was a very effective way of showing the different distributions. However, as the number of coins increases, the chart method becomes quite cumbersome. For example, in the case of 20 coins there are 1,048,576 different outcomes of which only 185,756 consists of 10 heads and 10 tails; while there 335,920 outcomes for which there are 11 of one kind and 9 of the other. next In fact the following chart may be of interest… next

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Heads and Tails Distribution of 20 coins. Distribution 20 heads ** 0 tails 19 heads ** 1 tail 18 heads ** 2 tails 17 heads ** 3 tails 16 heads ** 4 tails 15 heads ** 5 tails 14 heads ** 6 tails 13 heads ** 7 tails 12 heads ** 8 tails 11 heads ** 9 tails # of ways 1 20 190 1,140 4,845 15,504 38,760 77,520 125,970 167,960 9 heads ** 11 tails 8 heads ** 12 tails 7 heads ** 13 tails 6 heads ** 14 tails 5 heads ** 15 tails 4 heads ** 16 tails 3 heads ** 17 tails 2 heads ** 18 tails 1 heads ** 19 tails 0 heads ** 20 tails 1 20 190 1,140 4,845 15,504 38,760 77,520 125,970 167,96010 heads ** 10 tails 185,756 next Approximately, only 17.7% of the outcomes consist of 10 heads and 10 tails.

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Key Stone Problems… Key Stone Problems… next Set 1 © 2007 Herbert I. Gross.

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