# 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<-> (n-5)/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 5-digit 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. hypotenuse2 = 12 + 22 = 1 + 4 = 5, so hypotenuse = square root of 5.
b. If x is thee length of the other leg, then 12 + x2 = 32 , so 1 + x2 = 9. Then x2 = 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 h2 + 42 = 52, so h2 =52-42 = 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 = 105= 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:
 0-4 5-9 10-14 15-19 20-24 25-29 30-34 35-39 40-44 45-49 50-54 55-59 0 2 5 3 9 0 1 1 1 0 0 1

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.