key stone problem… key stone problem… next set 6 © 2007 herbert i. gross
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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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© 2007 Herbert I. Gross
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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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© 2007 Herbert I. Gross
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.
In exponent form 22 = 4.
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© 2007 Herbert I. Gross
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.
In exponent form 24 = 16.
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© 2007 Herbert I. Gross
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(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
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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.
In exponent form 25 = 32.
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© 2007 Herbert I. Gross
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(c) Twenty pennies are flipped simultaneously. In how manyways can the coins turn up
“heads" or “tails”?
Keystone Illustration for Lesson 5
Answer: There are 220 or 1,048,576 ways.
© 2007 Herbert I. Gross
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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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© 2007 Herbert I. Gross
Solution for Part c: For example…
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© 2007 Herbert I. Gross
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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© 2007 Herbert I. Gross
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
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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.
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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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© 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…
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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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© 2007 Herbert I. Gross
• 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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© 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 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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© 2007 Herbert I. Gross
• 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
© 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 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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© 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…
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© 2007 Herbert I. Gross
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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© 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.
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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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© 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, 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
© 2007 Herbert I. Gross
• 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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© 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.
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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.