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

Key Stone Problem…Key Stone Problem…

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Set 6© 2007 Herbert I. Gross

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

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© 2007 Herbert I. Gross

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.

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(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

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Answer: 16Solution 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.

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Solution for Part a:

• If the penny is flipped by itself, there are 2 possible outcomes.

In exponent form 21 = 2.

• If the penny and nickel are flipped by themselves there are 2 × 2 or 4 outcomes.


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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 23 = 8

• If the penny, nickel, dime, and quarter are flipped by themselves there are 2 × 8 or 16

outcomes.


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(b) How does the answer to part (a) change if a penny, nickel, dime, quarter,

and half dollar are flipped?.


Answer: The number of ways doubles. Therefore, there are 32 ways.


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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 16outcomes when the four coins were flipped.

Hence when the 5th coin is added, there will be 2 × 16 or 32 outcomes.


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(c) Twenty pennies are flipped simultaneously. In how manyways can the coins turn up

“heads" or “tails”?


Answer: There are 220 or 1,048,576 ways.


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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 2n possible outcomes.

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Solution for Part c: For example…

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Number of coins (n) Number of outcomes Exponential notation

1 2 21

2 4 22

3 8 23

4 16 24

5 32 25

6 64 26

7 128 27

Solution for Part c:

So with 20 coins the number of possible outcomes is 220, which is a far lesscumbersome way for expressing…

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2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

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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.

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• 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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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…

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n 2n n 2n

1 2 11 2,0482 4 12 4,0963 8 13 8,1924 16 14 16,3845 32 15 32,7686 64 16 65,5367 128 17 131,0728 256 18 262,1349 512 19 524,288

10 1,024 20 1,048,576

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• Mathematically, an answer such as 220 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 xy key we simply use the following sequence of key strokes…

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2 xy 20 =

and the answer 1,048,576 appears in the display window.

1,048,576

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• 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 25 = 32. If we look at 210 in its expanded form; that is, as…

2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

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by the associative property we may rewrite it as…

(2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2)

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• We know from our chart that 25 = 32. Hence we may rewrite …

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( 2 × 2 × 2 × 2 × 2 ) × ( 2 × 2 × 2 × 2 × 2 )4 48 816 1632 32 = 1,024as

• 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


• 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 Nickel Penny

Outcome #3

Outcome #4

Outcome #5

Outcome #6

Outcome #7

Outcome #8

heads

heads

heads

heads

tails

tails

tails

tail

heads

tails

heads

tails

heads

heads

tails

tails

heads

tails

heads

tails

heads

tails

heads

tails

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• 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…

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dime nickel penny quarterheads heads heads headsheads heads heads tailsheads heads tails headsheads heads tails tailsheads tails heads heads

Outcome 1Outcome 2Outcome 3Outcome 4Outcome 5Outcome 6Outcome 7Outcome 8Outcome 9Outcome 10Outcome 11Outcome 12Outcome 13Outcome 14Outcome 15Outcome 16

heads tails heads tailsheads tails tails headsheads tails tails tailstails heads heads headstails heads heads tailstails heads tails headstails heads tails tailstails tails heads headstails tails heads tailstails tails tails headstails tails tails tails

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• 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.

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• 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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• 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, thereare 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…

Distribution Number of Ways4 heads, 0 tails 13 heads, 1 tails 42 heads, 2 tails 61 heads, 3 tails 40 heads, 1 tails 1Total Outcomes 16


• Hence 3 of one kind and 1 of the other will appear 8 times…

dime nickel penny quarterheads heads heads headsheads heads heads tailsheads heads tails headsheads heads tails tailsheads tails heads heads

Outcome 1Outcome 2Outcome 3Outcome 4Outcome 5Outcome 6Outcome 7Outcome 8Outcome 9

Outcome 10Outcome 11Outcome 12Outcome 13Outcome 14Outcome 15Outcome 16

heads tails heads tailsheads tails tails headsheads tails tails tailstails heads heads headstails heads heads tailstails heads tails headstails heads tails tailstails tails heads headstails tails heads tailstails tails tails headstails tails tails tails

and 2 of one kind and 2 of the other will appear only 6 times.

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• 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.

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• In fact the following chart may be of interest…

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Heads and TailsDistribution

of

20 coins.

Distribution20 heads ** 0 tails19 heads ** 1 tail 18 heads ** 2 tails17 heads ** 3 tails16 heads ** 4 tails15 heads ** 5 tails14 heads ** 6 tails13 heads ** 7 tails12 heads ** 8 tails11 heads ** 9 tails

# of ways1

20190

1,1404,845

15,50438,76077,520125,970167,960

9 heads ** 11 tails8 heads ** 12 tails7 heads ** 13 tails6 heads ** 14 tails5 heads ** 15 tails4 heads ** 16 tails3 heads ** 17 tails2 heads ** 18 tails1 heads ** 19 tails0 heads ** 20 tails 1

20190

1,1404,845

15,50438,76077,520

125,970167,960

10 heads ** 10 tails 185,756

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Approximately, only 17.7% of the outcomes consist of 10 heads and

10 tails.

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

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