M 302, Spring 05 (Smith)
SOLUTIONS TO QUIZ PROBLEMS
1/28: The total profit was $2: On the first round, the dealer
paid out $7 and made $8, for a profit of $1. On the second round, the
dealer paid out $9 and made $10, again for a profit of $1. So the total
profit was $1 from first round + $1 from second round = $2.
Another method of explanation: If the dealer started with $10, then
after buying for $7, she had $3 left. After selling it for $8, she had
$3 + $8 = $11. After buying it back for $9, she had $11  $9 = $2.
Finally, when she sold it for $10, she had $2 + $10 = $12, so her total
profit was $12  $10 = $2.
1/31: First problem: If there are 18 mangoes in each of 5
layers, then there are 5 X 18 = 90 mangoes in all, so this is a good
estimate.
Second problem: No: If each shelf can contain at most three CD's, then
the most you could have on the five shelves altogether is 5 X 3
= 15 CD's. So since you have 16 CD's in all, you will still have one
CD left over.
2/4: a. You need 3 socks: If you just have two, they might have
different colors. With 3 socks, there are two cases:
Case 1: The first two socks have the same color. Then they
make the desired pair
Case 2: The first two socks are different colors. Then the third
sock must match one or the other of the first two, since there are only
two colors possible.
b. You need 5 socks: With 4 socks, you might have 3
of one color and 1 of the other, so would have only one pair. With
five socks, there are two cases:
Case I: There are at least 2 socks of each color. Then you
have the desired two pairs.
Case II: There is one color for which you only have one sock. Then the
remaining 4 socks must all be the other color, so you
have two pairs of that color.
c. You need 12 socks: If you have 11 socks, you
might have 10 blue and 1 black. But if you have 12, then since there
are only 10 blues socks available, you must have at least 2 black socks.
2/7: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610
2/9: Start by finding the largest Fibonacci number less than
609. That's 377. Subtract: 609  377 = 232. (So 609 = 377 + 232.) Now
find the largest Fibonacci number less than 232 and subtract: 232  144
= 88. (So 232 = 144 + 88, so 609 = 377 + 144 + 88.) Continue this
process: 88 
55 = 33, 33  21 = 12, 12  8 = 4, 4  3 = 1. Putting it all together:
609 = 377 + 144 + 55 + 21 + 8 + 3 + 1.
2/11: a. 10:45, since 96 hours is exactly 4 days (4 x 24 = 96).
b. 1063 divided by 12 is 88, with remainder 7. In 88
x 12
= 1056 hours, it will be 10:45 again, so 7 hours later (1063 hours from
now)
it will be 5:45.
c. 24 hours ago, the clock read 7:10. But 23 hours ago was one hour
later than 24 hours ago, so the clock read 8:10 then.
2/14: a. 16 divided by 7 gives remainder 2, so 16 is equivalent
to 2 (mod 7).
b. 24 divided by 7 gives
remainder 3, so 24 is equivalent to 3 (mod 7).
c. Using the results of the
previous parts, 16 x 24 is equivalent to 2 x 3 = 6 (mod 7).
2/16: Please Note: I copied the third number incorrectly
on the quiz, so only the first two parts will be graded.
a. First add the digits in the odd places mod 10:
0 + 1 + 0 + 0 + 5 + 2 is equivalent to 8 (mod 10)
Multiply that result by 3: 3 x
8 = 24 is equivalent to 4 (mod 10)
Add the digits in the even places
mod 10: 5 + 0 + 0 + 2 + 6 + 4 = 17 is equivalent to 7 (mod 10).
Add the last two results mod 10:
4 + 7 = 11 is equivalent to 1 mod 10.
Since the final result is not 0
mod 10, this is not a valid UPC.
b. First add the digits in the odd places mod 10:
0 + 1 + 0 + 0 + 5 + 6 = 12 is equivalent to 2 (mod 10)
Multiply that result by 3: 3 x 2
= 6 (mod 10)
Add the digits in the even places
mod 10: 5 + 0 + 0 + 2 + 2 + 4 = 13 is equivalent to 3 (mod 10).
Add the last two results mod 10:
6 + 4 = 9 mod 10.
Since the final result is not
0 mod 10, this is also not a valid UPC.
2/21:
a. Since 125 = 5 x 25 and 10 = 5 x 2, we know 125/10
= 25/2. Since 25 and 2 have no common factors, this is in lowest terms.
b. We can multiply fractions by multiplying
numerators and multiplying denominators to get the numerator (top
number) and denominator (top number) of the product, so (1/2) x (6/5) =
(1 x 6)/(2 x 5) = 6/10 (which may be reduced to 3/5, if you wish).
c. A couple of possible method:
I. 168.5 = 
(168 + 1/2) =  (336/2 + 1/2) = (337/2) = (337)/2
II. 168.5 =
(1685/10) = (1685)/10
2/23: a. 123.123123... b.
4.999...
c. 0.986 d.
0.0333...
2/25: a. 1.289901 = (1.28902 x 100,000)÷(100,000) =
(128,901)/(100,000)
b. If x = 2.222..., then
10x = 22.222...
x = 2.222... Performing the subtraction gives:
9x = 20, so x = 20/9.
2/28: a. There are many possible answers. Some examples:
12.03456911, 12.03456912, 12.03456913, 12.034569115, 12.0345691102, etc.
b. There are many possible answers. One that uses ideas discussed in
class: 5.7101001000100001.... (Since the pattern continues indefinitely
but never repeats, this number must be irrational. If you just said, "a
pattern that does not repeat and will not terminate," without giving an
explicit example, you only got partial credit.)
3/2: a. There is more than one way to do this, but perhaps the
easiest: The number n in E is paired with the number n  1 in O.
In other
words, 2<>1, 4<>3, 4<>5, etc.
b. The easiest pairing: 5<>1, 15<>2, 25<>3, etc.
Describing this in words is a little tricky: Take a number in EIF
and pair it with the number you get by dropping the 5 at the end, then
treating the result as a number with one less digit and adding 1. In
symbols, it's even trickier: n<> (n5)/10 + 1. Going the other
way is a little easier: the natural number m corresponds to 10(m+1) +
5, which is the number you get by putting an extra digit 5 at the end
of m + 1.
3/4: a. There are lots of possible answers  for example,
12345,
or any other 5digit number that does not have a 3 in the first place,
does
not have a 5 in the second place, and does not have a 2 in the third
place.
b. No  since you don't know any of the digits in
the
third number, there is no way you could know whether or not the number
you
describe would be different from this third number.
3/7: a. hypotenuse^{2} = 1^{2} + 2^{2} =
1
+ 4 = 5, so hypotenuse = square root of 5.
b. If x is thee length of the other leg, then 1^{2}
+ x^{2} = 3^{2} , so 1 + x^{2} = 9. Then x^{2}
= 9  1 = 8, so x = square root of 8.
3/9: (Sorry, I'm not able to draw the picture here.) Draw your
own
picture of a rectangle with one diagonal drawn in. Label the base 4",
the
height h, and the diagonal 5". In your diagram you will see a triangle
with
base 4", height h, and hypotenuse 5". Applying the Pythagorean Theorem
to
this triangle, we have h^{2} + 4^{2} = 5^{2},
so h^{2}
=5^{2}4^{2} = 25  16 = 9, giving h = 3. Then we can
find
the area of the rectangle as A = h X 4 = 3 X 4 = 12 sq. inches.
3/21:

Number of
faces

Number of
edges

Number of
vertices

Cube

6

12

8

Tetrahedron

4

6

4

3/23: a. There are 100  53 =
47 heads, so the proportion of heads is 47/100 = .47, and the
percentage is therefore 47%.
b. The proportion is 28/50 =
56/100 = .56, so the percentage is 56%.
c. 100%  56% = 44%. (Or you
could reason: 50  28 = 22, so the proportion of heads is 22/50 =
44/100 = .44, giving percentage 44%.)
3/28: a. There are various ways
to list the outcomes. One possibility:
1. Penny H, dime H
2. Penny H, dime T
3. Penny T, dime H
4. Penny T, dime T
b. Possibilities 1, 2, 3 from part (a).
c. 3/4, since E occurs in three of the four equally likely
possibilities.
3/30: We are randomly choosing
two people and looking at their birthdays. So the possible outcomes are
the possible pairs of birthdays. (We must consider ordered pairs, since
we are distinguishing between the first person's birthday and the
second person's birthday.) Assuming there are 365 equally likely
birthdays, we get 365 x 365 equally likely outcomes. There is only one
outcome, namely the pair of birthdays (Dec. 9, Dec. 9), that
corresponds to the event we are trying to find the probability of
(namely, that both people are born on December 9). So the probability
that both people are born on December 9 is 1/(365 x 365), which is
about 0.0000075  extremely small!
4/4: a.

1

2

3

4

5

6

H

H1

H2

H3

H4

H5

H6

T

T1

T2

T3

T4

T5

T6

b. There are twelve possible outcomes
shown above. Only one of these (H4) fits the conditions of getting a
head on the coin and a 4 on the die. So the probability of getting a
head on the coin and a 4 on the die is one out of twelve, or 1/12.
c. Once we know that the die shows a 2, we only have two possibilities:
H2 and T2. Only one of these has tails on the dime, so the probability
of the dime showing tails when we know that the die shows a 2 is one
out of two, or 1/2.
4/6: a. There are 10 choices (0
through 9) for each of the five digits, so the total number of codes is
10×10×10×10×10 = 10^{5}= 10,000.
b. Since there are (from part a) 10,000 possible
outcomes for the random guess, and 200 of these will open the door (one
for each of the 200 students), the probability that a random guess will
open the door is 200/10,000 = 1/500 = .002.
4/6: a. If we don't care about
order, the number of lists of 3 cards is (#choices for first
card)x((#choices for second card)x(#choices for third card) = 52x51x50
(= 132600). But each of these is counted 6 times: (#choices for first
of 3 cards in the list)x(#choices for second card in list)x(#choices
for third card in list) = 3x2x1=6. So we need to divide the number of
ordered lists by the number of times each set is listed, to get answer
(52x51x50)/(3x2x1) (= 132600/6 = 22100).
b. Again, we are interested in the number of sets of
things, where order does not matter. Since we have 40 numbers to choose
from and are choosing 6 different numbers, the same reasoning as above
shows that the number of unordered sets of 6 numbers chosen from 40 is
(40x39x38x37x36x35)/(6x5x4x3x2x10) ( = 27,636,600/720 = 3,838,300).
Finally, the probability of winning is one outcome out of this number,
or 1/3,838,300 (also expressible as (6x5x4x3x2x10)/(40x39x38x37x36x35),
or about .0000002605)
4/8: a. We can calculate the
number of ordered lists of 3
cards as (52 choices for the first card) x (51 choices left for the
second card) x (50 choices left for the third card) = 52x51x50 =
132600. But we want the number of unordered
groups of 3 cards. So we need to figure out how many ways each
unordered group of 3 cards has been counted in the ordered lists, and
divide the total number of ordered lists by that number. Well, if we
consider how many ways we can order the same 3 cards, we have (3
choices for the first card) x (two choices left for the second card) x
(only 1 choice left for the third card) = 3 x 2 x 1 = 6 possible ways
of ordering the three cards. So the number of ordered lists is
(52x51x50)/(3x2x1) = 132600/6 = 22100.
b. By the same kind of reasoning as in part (a), the
number of possible collections of 6 different numbers you can pick from4/11: a total of 40 numbers is
(40x39x38x37x36x35)/(6x5x4x3x2x1) = 2763633600/720 = 3838300, so the
probability of winning is the probability of picking the one winning
combination out of this many: 1/3838300 = 0.00000026
4/11: a. Since it's reasonable
to assume that whether or not the teacher brings donuts is independent
of whether or not it is sunny, we can calculate:
probability that it will be sunny and the teacher will bring
donuts = (probability that it will be sunny)x(probability that the
teacher will bring donuts) = (1/2)x(1/10) = 1/20
b. The number of ways of choosing 3 cards from 52 is
(52 ways to choose first card)x(51 ways left to choose second card)x(50
ways left to choose third card) = 52x51x50 = 132600. The number of ways
of choosing 3 jacks from a total of 4 jacks in the deck is (4 ways of
choosing first jack)x(3 ways left to choose second jack)x(2 ways left
to choose third jack) = 4x3x2 = 24. So the probability of choosing 3
jacks is 24/132600 = 0.000181.
4/13: We expect about 50 of the
students will get heads, and all of these will say yes. That leaves an
estimated 54  50 = 4 students who said yes but got tails  so our
estimate is that 4 out of 50 (the 50 who got tails) said yes because
they have cheated on an exam. So the estimated fraction who cheated is
4/50, giving estimate of (4/50)x100 out of 100 who have cheated.
4/15: Since there are 9 items
altogether (the burger can go in any order, too), there are 9 choices
for the item that goes on first, 8 for the item that goes on second
(since each item goes on only once), 7 for the item that goes on third,
and so on, until there is just one choice left for the item that goes
on last. This gives 9x8x7x6x5x4x3x2x1 possible orderings (that is,
possible "burgers with the works") in all.
4/18: a. The mean is (72 + 84 +
61 + 95 + 92+ 98 + 87 + 84)/8 = 673/8 = 84.125
b. To find the median, first put the numbers in
order: 61, 72, 84, 84, 87, 92, 95, 98. Since we have an even number of
numbers in the list, the median is the average of the middle two: (48 +
87)/2 = 85.5.
c. Since 80 is lower than the mean, adding a score
of 80 will lower the mean. To check this out: (72 + 84 + 61 + 95 + 92+
98 + 87 + 84 + 80)/9 = 83.67, which is less than 84.125.
d. In this case, the new score lowers the median as
well: The new set of scores in order is 61, 72, 80, 84, 84, 87, 92, 95,
98. Since there is now an odd number of scores, the median is the
middle number, 84, which is lower than the earlier median of 85.5.
(Sometimes the median will not be lowered by adding a number that is
less than the median. For example, if the original scores were 80, 84,
84, 84, 88 and we added a new score of 80, the old and new median would
both be 84.)
4/20: a. First, it's helpful to
make a chart of how many of the numbers are in each interval:
04

59

1014

1519

2024

2529

3034

3539

4044

4549

5054

5559

0

2

5

3

9

0

1

1

1

0

0

1

Click
here to see the histogram.
b. The distribution of the data can be described as: concentrated
between 5 and 25, especially between 20 and 25, and then a dribble off
to the right. ("Skewed to the right" is the technical term.)
4/22: Since the numbers are
already listed in order, it is relatively easy to find the five number
summary:
The min and max are the smallest (first) and largest (last) numbers in
the list: 5 and 58.
The median is the middle number in the list. Since there are 23 numbers
in the list, the median is the 12th from the top (or the 12th from the
bottom): 20.
The quartiles are the medians of the lower and upper halves: Since
there are 11 numbers above the median and 11 below, we can take the
quartiles to be the middle numbers of these groups of 11  that is,
the sixth number from the bottom (12) is the lower quartile, and the
sixth number from the top (21) is the upper quartile.
4/25: a. The mean is
(72+93+88+85+97+100+77)/7 = 87.42, about 87. So the distances from the
mean are (respectively) about 15, 6, 1, 2, 10, 13, 10. Only two of
these are near 2.7; the others are much larger. So 2.7 isn't a "typical
distance from the mean." 10.3 fits this description better  indeed,
the median of the differences is 10, very close to 10.3. (Note: The
actual calculations will give a standard deviation of 10.277.)
b. There are many possible correct answers. One is:
0, 3, 3. The median of these three numbers is 3, but the mean is
(0+3+3)/3 = 2, lower than the median.
4/27: Both are examples of post hoc, ergo propter hoc.
In (a), just because the penny was picked up before the person bought
the lottery ticket doesn't mean that picking up the penny caused the
win. There are lots of other things the person did before buying the
ticket, but there is no way any of them could influence the outcome of
the lottery.
In (b), there is no evidence that the dog lick must have caused the illness. The
dog might have caused the
illness, but there are lots of other things that could have caused it
as well.
4/29: 1. The expected value is
(probability of getting $1)x($1) = (probability of getting $.50)x($.50)
= (1/2)x($1) + (1/2)x($.50) = $.50 + $.25 = $.75.
2. a. 1/2 (red is half of the spinner).
b. 1/4 the spinner is blue
and 1/8 is green, so the probability of getting blue or green is 1/4 +
1/8 = 3/8.
5/1: The probability of winning
is 18/38. The probability of losing is 20/38. Call the payoff P. Then
the expected value of the bet is
(probability of winning)x(net gain) + (probability
of losing)x(loss)
= (18/38)(P$1000) + (20/38)($1000).
This needs to be zero for the game to be fair. So we need to solve the
equation
(18/38)(P$1000) + (20/38)($1000) = 0.
To do this, first multiply both sides of the equation by 38:
(18)(P$1000) + (20)($1000) = 0.
Now multiply out each part:
18P  $18000  $20,000 = 0.
Collect the dollar amounts:
18P  $38,000 = 0.
Finally,
18P = $38,000
P = $38,000/38 = $2111.11.