Attached.

Part A: Questions 1 – 8

1.
a. Law of Large Numbers
Ans. In statistics, the law of large numbers is the result of many trials of the same
experiment. As the number of trials increases, the average results must be approximately or
equal to the expected value. The best example is the flipping of the coin. Since the coin has
two sides, (heads and tails) it is safe to assume that the theoretical probability of getting the
heads or tails is 50 percent or 0.5. If you apply the law of large numbers, flipping the coins 10
or 20 times does not guarantee that you will get 10 heads or tails. However, if you flip the
coins indefinitely, the cumulative proportions of head or tails will approach approximately
equal to 50%.

b. The non-existent law of averages
Ans. Basically, the law of averages is just belief that means small numbers also apply to
large numbers. Unlike the Law of Large numbers, the law of averages does not contain
statistical concepts. For example, you and your friends are playing a color game. Then, one
of your friends says you should bet on color red will come out because after several rolls,
there’s still no red color. This is obviously not true because of its uncertainty and it needs
statistical analysis to have a higher chance of getting the red color.

c. Fundamental Counting Principle (for “and” & “or”)
Ans. If event A occurs in x ways or event B occurs in y ways, then A OR B can occur in x+y
ways. Meaning if you see a problem with “or”, you must use addition. For instance, if an
event A occur in x ways and event B occurs in y ways, then A AND B can occur in x*y ways.
In short, if “and” is used, you must use multiplication. This can also be observed in the truth
table of “AND” and “OR”.

d. Permutation
Ans. Permutation is commonly defined as an ordered arrangement of a finite number of the
elements; either all the available “n” elements or of a part of them. One of the best examples
of getting permutations is getting how many ways three people win the first place, second
place and third place. This is a question about permutation because it has an ordered
arrangement.
The formula for permutation is 𝑃(𝑛, 𝑟) =

𝑛!
(𝑛−𝑟)!

where P is the number of permutations
n is the total number of objects
r is the number of choosing objects from the set

e. Combination

Ans. Combination is defined as an arrangement of the selection of objects regardless of the
order. For example, there are four balls of different colors. Then, two balls are taken and
arranged in any way. This is a combination because there is no specific order in the results.
The formula for combination is 𝐶(𝑛, 𝑟) =

𝑛!
𝑟!×(𝑛−𝑟)!

where C is the number of combinations
n is the total number of objects in the set
r is the total number of choosing objects from the set

2.
Ans. There are many players in a raffle. You can get 1 prize from the 4 prizes, but can you
win against all the people who entered the raffle. Think of it this way, I joined a raffle of 5000
people competing for one of the four prizes. I know I can only get one prize, since if you are
already picked, you will be removed from the raffle. My chance of winning is 4 out of 5000,
since only 4 people can win out of 5000 who entered the raffle.

3.
a.
Ans. Including me, there will be 19 girls in this problem. Using the formula for combination
10!
19!
𝐶(𝑛, 𝑟) = 𝐶(10,1) ∗ 𝐶(19,1) =
×
= 190 𝑤𝑎𝑦𝑠. Multiplication operator was
1! ×(10−1)!

used because of the word “AND”.

1!(19−1)!

*Note. If you are a male, make 11 as the number of boys, and 18 for girls and use the same formula.

b.
Ans. 𝐶(𝑛, 𝑟) = 𝐶(10,1) ∗ 𝐶(19,1) =
used because of the word “OR”.

10!
1! ×(10−1)!

+

19!
1!(19−1)!

= 29 𝑤𝑎𝑦𝑠. Addition operator was

4.
Ans. Using the formula for permutation, 𝑃(𝑛, 𝑟) = 𝑃(12,4) =

12!
(12−4)!

= 11880 𝑤𝑎𝑦𝑠

*Note. When you see arranged or when there is an arrangement in picking, use permutation. When
someone just needs to pick something regardless of order/arrangement. Use combination

5.
a.
Ans. We use combination since the order does not matter.
𝐶(𝑛, 𝑟) = 𝐶(8,3) × 𝐶(6,2) =

8!
6!
×
= 840
3! (8 − 3)! 2! (6 − 2)!

Note* That whenever you see the word “AND” it always means multiply the different ways, and when you
see, OR” it means add. You will learn this when you go to Logic systems in school.

b.
Ans. We will use the principle of probability to answer this question. 𝑃 =
𝑃(𝐽𝑎𝑘𝑒 𝑜𝑛 𝑐𝑜𝑚𝑚𝑖𝑡𝑡𝑒𝑒) =

𝑛𝑜 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠
𝑡𝑜𝑡𝑎𝑙 𝑛𝑜 𝑜𝑓 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠

𝑛𝑜 𝑜𝑓 𝑤𝑎𝑦𝑠 𝐽𝑎𝑘𝑒 𝑐𝑎𝑛 𝑏𝑒 𝑐ℎ𝑜𝑠𝑒𝑛
𝑛𝑜 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑚𝑚𝑖𝑡𝑡𝑒𝑒𝑠 𝑤𝑖𝑡ℎ 3 𝑓𝑟𝑒𝑠ℎ𝑚𝑎𝑛 𝑎𝑛𝑑 2 𝑠𝑜𝑝ℎ𝑜𝑚𝑜𝑟𝑒

𝑃(𝐽𝑎𝑘𝑒 𝑜𝑛 𝑐𝑜𝑚𝑚𝑖𝑡𝑡𝑒𝑒) =

𝐶(1,1)×𝐶(7,2)×𝐶(6,2)

=0.375

840

Let’s breakdown the numerator first, 𝐶(1,1) is Jake being chosen. Then 𝐶(7,2) is picking 2
more freshmen to be part of the committee (not three because Jake is already chosen. Then
𝐶(6,2) is picking 2 sophomores for the committee.
Then the denominator is just how many ways there will be 3 freshmen and 2 sophomores,
which we solve in letter a.
We are basically dividing the outcome the Jake will chose for a committee versus the real
outcome to get the probability of Jake will be chosen for a committee.

6.
a.
Ans. We use combination here since the arrangement of the cards does not matter. Like if
you got a six of diamonds and a 5 of spades. It doesn’t matter what card got into your hand
first, but what matters is the cards in your hand after picking.
52C5, that is 52 cards, and you are getting 5 cards, 𝐶(52,5)

52!
= 2598960 ways
5!(52−5)!

b.
Ans. For this question, we have 52 cards, 4 cards are aces. So, let’s remove that from the 52
cards. It is 4𝐶3. Now, we are left with 48 cards, where it can be any card, because it is a filler
card. We only need three aces. 48𝐶2, which means we will get only two cards from the deck.
This question is an “and” question. How many hands can I get with 3 aces and 2 other cards. So, the
formula is
4C3 *48C2 = 4512 𝐶(4,3) × 𝐶(48,2) =

4!
48!
×
3!×(4−3)!
2!×(48−2)!

= 4512 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠

c.
Ans. We see that “or” so we know its multiplication, the thing is, you need to look at this from
two separate parts, segregate the statements. We need 1 ace and 2 aces. So, let us solve
for the 1 ace first.
For the 1 ace, it is 𝐶(4,1) × 𝐶(48,4) =

4!
1!(4−1)!

×

48!

. From the 4 aces, we get 1. And from

4!(48−4)!

the 48 cards left, we get 4.
For the 2 aces, it is 𝐶(4,2) × 𝐶(48,3) =
from the 48 cards left, we get 3.

4!
2!(4−2)!

×

48!

. From the 4 aces we get two, and

3!(48−3)!

So the whole equation is

4!
1!(4−1)!

×

48!
4!(48−4)!

+

4!
2!(4−2)!

×

48!
3!(48−3)!

= 882096 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑤𝑎𝑦𝑠 . That

addition sign is the “or” word in the sentence.

7.
a.
Ans. 26*26*10*10*4 = 338000 possible ways.
Note: 26 is the total number of alphabets, then 10 is the total number of numbers from (0-9).
4 is the total number of characters used in the problem.

b.
Ans. 𝑃(𝐴𝑂52$) =

1×1×10×10×4
338000

=

400
338000

=

1
845

≈ 0.00118

8.
Ans. We will use the principle of probability.
𝑃 (𝑛𝑜 ℎ𝑖𝑠𝑝𝑎𝑛𝑖𝑐𝑠) =

𝑛𝑜 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 𝑤𝑖𝑡ℎ 𝑛𝑜 ℎ𝑖𝑠𝑝𝑎𝑛𝑖𝑐𝑠
𝑛𝑜 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑤𝑖𝑛𝑛𝑒𝑟𝑠

16!
𝐶(16,6) 6! (16 − 6)!
=
=
= 0.1073
22!
𝐶(22,6)
6! (22 − 6!)
Even though the probability is low, there will still be winners in Hispanics. However, it may
still occur (no Hispanic winners) due to random chance.

Part B: Questions 9 – 12

9.
Ans. 𝑃(𝑚𝑎𝑙𝑒 ∩ 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 ) = 𝑃(𝑚𝑎𝑙𝑒) × 𝑃(𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 |𝑚𝑎𝑙𝑒)
𝑃(𝑚𝑎𝑙𝑒 ∩ 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 = (0.36)(0.85)
= 0.306 𝑜𝑟 30.6% 𝑜𝑓 𝑚𝑎𝑙𝑒 𝑖𝑠 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 𝑤𝑖𝑡ℎ 𝑡ℎ𝑒 𝑐𝑎𝑟𝑒 ℎ𝑒 𝑖𝑠 𝑟𝑒𝑐𝑒𝑖𝑣𝑖𝑛𝑔

10.
a. Draw a Venn diagram.
Ans.

Bill B

Bill A

0.35 – 0.14 =
0.21

0.14

0.62 – 0.14 =
0.48

U = 1 – 0.21 – 0.48 – 0.14 = 0.17

b.
i.
In Venn diagram, “OR” means union. So, the union of Bill A and Bill
B is 𝐵𝐴 ∪ 𝐵𝐵 = 0.21 + 0.14 + 0.48 = 0.83
ii.
Not means negation. In the Venn diagram, Bill A must not be
included.
𝐵𝐵 = 0.48
iii.
This means that Bill A and Bill must not be included.
𝑈 = 0.17
11. While looking over the test scores for one of her classes, a teacher tabulated that of
the 24 students, 17 students completed all the homework for the unit, 18 students
passed the unit test, and 15 students completed all the unit homework and passed the
test. (16 pts)

a.
Ans. (1) Put the Total = 24 in Row 4 Column 4.
(2) Put 17(given) in Row 2 Column 4.
(3) Put 15(given) in Row 2 Column 2.
(4) Subtract 17-15 to get 2. Put it in Row 2 Column 3.
(5) Put 18(given) in Row 4 Column 2.
(6) Subtract 18-15 to get 3. Put it in Row 3 Column 2.
(7) Subtract 24-17 to get 7. Put it in Row 3 Column 4.
(7) Subtract 7-3 to get 4. Put it in Row 3 Column 3.
(8) Add 2+4 to get 6. Put it in Row 4 Column 3.
Passed
Did Not Pass
Completed
15
2
Did Not Complete
3
4
Total
18
6

Total
17
7
24

b.
i.
Ans.

18
24

= 0.75

ii.
Ans.

7

24

≈ 0.2917

iii.
Ans.

15
24

= 0.625

iv.
15

Ans.

17

≈ 0.8824

v.
15

Ans.

18

≈ 0.8333

c.
Ans. In this example, the test success is not independent in the homework
completion because P(success) = 0.6087 while the P(success|completed) = 0.6667.
If these two are independent, they should have the same probabilities.

12.
a.
Ans.

13
22

14
23

Black P (black and black) = 0.3587

black
9
22

14
22

9
23

Black P (black and orange) = 0.2490

Black P (orange and black) = 0.2490

orange
6
22

Black P (orange and orange) = 0.1423

b.
Ans. Using the tree diagram above, the probability that his second pick is an
9
orange candy is ≈ 0.4091
22

c.
i.
Ans. Based on the diagram above, the probability that Ramon picks
9

14

63

an orange candy then black candy is (23) (22) = 253 ≈ 0.2490
ii.
Ans. The probability that Ramon picks a black candy given that he
14
has already picked and orange candy is ≈ 0.6364
22

iii.
Ans. The probability that Ramon picks two different colored cadies
is 253 + 253 = 253 ≈ 0.4980
63

63

126

iv.
Ans. The probability that Ramon picks no orange candies is
14

13

91

(23) (22) = 253 ≈ 0.3597

Part C: Questions 13 – 15

13.
a.
Ans. For this problem, i...