Practice Test 2A Probability Solution

1. 1
PA B  0
4. The probability of an event is always between 0 and 1
0 ≤ P(A) ≤1
5. The sum of probabilities of all final outcomes of an experiment is always 1
6. The number of Combinations of n items selected n at a time is 1
Statistics: Practice Test 2A (Part One) Solution
Probability
1. A simple event is an outcome or event that cannot be further broken down.
2. A sample space is a procedure that consists of all possible sample events.
3. If two events are mutually exclusive, the probability that both will occur is 0
nCr =
𝑛!
(𝑛−𝑟)!𝑟!
nCn =
𝑛!
(𝑛−𝑛)!𝑛!
=
𝑛!
0!𝑛!
=
𝑛!
𝑛!
= 1
7. The number of Permutations of n items selected 0 at a time is 1
nPr =
𝑛!
(𝑛−𝑟)!
nPo =
𝑛!
(𝑛−𝑜)!
=
𝑛!
𝑛!
= 1
8. The sample space for tossing 5 coin s consists of how many outcomes? 32 outcomes
each coin has 2 outcomes: 25
= 32
Since there are only 2 possible outcome for each coin, Tail (T) or Head (H), and
there are 5 coins, Number of outcomes will be = 2 x 2 x 2 x 2 x 2 = 25 = 32
In general: Size of Sample Space = (# of outcomes per stage) # of stages
Sample Space:
{HHHHH, TTTTT, HHHHT, HHHTH, HHTHH, HTHHH, THHHH, TTTTH, TTTHT,
TTHTT, THTTT, HTTTT, HHHTT, HHTTH, HTTHH, TTHHH, HHTHT, HTHHT,
THHHT, HTHTH, THTHH, THHTH, TTTHH, TTHHT, THHTT, HHTTT, TTHTH,
THTTH, HTTTH, THTHT, HTHTT, HTTHT}
9. A quiz consists of 3 true-false questions, how many possible answer keys are there? 8
possible answer keys
23
= 8
Sample Space:
{TTT, FFF, TFF, FTF, FFT, FTT, TFT, TTF}
10. A pizza parlor offers 10 different toppings; how many four topping pizzas (different
toppings) are possible? 210 four topping pizzas are possible

2. 2
order is not important  combination is used
nCr =
𝑛!
(𝑛−𝑟)!𝑟!
10C4 =
10!
(10−4)!4!
=
10𝑥9𝑥8𝑥7
4𝑥3𝑥2
= 210
11. How many 6-letter code words can be made from the 26 letters of the alphabet if no letter
can be used more than once in the code word? 165, 765, 600 6-letter code words can be
made
order is important, no letter can be used more than once  permutation is used
26P6 =
26!
(26−6)!
=
26𝑥25𝑥24𝑥23𝑥22𝑥21
1
= 165,765, 600
12. A random sample of 100 people was asked if they were for or against the tax increase on
rich people. Of 60 males 45 were in favor, of all females 22 were in favor. Write the
contingency table and answer the following questions. (Hint: Make up a table like Table
4-1 of page 152.) If one person is selected at random, find the probability that:
For tax increase on rich people
(T)
Against tax increase on rich people
(A)
Total
Male (M) 45 15 60
Female
(F)
22 18 40
Total 67 33 100
a) This person favors the tax increase on rich people.
P(T) = n(T) / n(S)
P(T) = 67/100 (or 0.67)
b) This person is a female.
P(F) = n(F) / n(S)
P(F) = 40/100 = 2/5 (or 0.4)
c) This person opposes the tax increase on rich people given that the person is a female.
P(A|F) = P(A ∩ F) / P(F)
P (A|F) = n(A ∩ F) / n(F)
P(A |F)= 18/40 = 9/20 (or 0.45)
d) This person is a male given that he favors the tax increase on rich people.
P(M|T) = P(M ∩ T) / P(T)
P(M|T) = n(M ∩ T) / n(T)
P(M|T) = 45/67 (or 0.6716)
e) This person is a female and favors the tax increase on rich people.
P (F∩ T) = 22/100 = 11/50 (or 0.22)
f) This person opposes the tax increase on rich people or is a female.
P(A U F) = P(A) + P(F) - P(A∩F)

3. 3
P(A U F) = n(A)/n(S) + n(F)/n(S) – n(A∩F)/n(S)
P(A U F) = 33/100 + 40/100 – 18/100 = 55/100 = 11/20 (or 0.55)
g) Are the events “females” and opposes the tax increase on rich people independent?
Explain.
Events “females” and opposes the tax increase on rich people is dependent because
occurrence of one of the events does affect the other event.
Proof: P (A |F) = 18/40 ≠ P (A) = 33/100
h) Are they mutually exclusive? Explain.
They are not mutually exclusive because a person can be both female and oppose the
tax increase on rich people. Those two events are not disjoint.
Proof: P (A∩F) = 18/100 ≠ 0
13. Find the probability of getting the outcome of “Tails and 2” when a coin is tossed and a
die is rolled.
P(A∩B) = P(A) x P(B):
𝑃(𝑇𝑎𝑖𝑙𝑠 𝑎𝑛𝑑 2) = 𝑃(𝑇𝑎𝑖𝑙𝑠) × 𝑃(2) =
1
2
×
1
6
=
1
12
= 0.0833
14. A box consists of 14 red and 36 blue markers. If we select 3 different markers randomly,
a. What is the probability that they are all red? (with replacement)
n(total) = 14 + 36 = 50
P(R) = n(R)/n(S) = 14/50 = 7/25
𝑃(1 𝑟𝑒𝑑) =
𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑑
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑟𝑘𝑒𝑟𝑠
=
14
14+36
=
14
50
= 0.28
P(3 red) = 0.283
= 0.021952
b. What is the probability that they are all red? (without a replacement) Draw a tree
diagram and label each branch.
P(all 3 red) =
14
50
×
13
49
×
12
48
=
2184
117600
=
13
700
= 0.0186

4. 4
15. If the probability of winning the race is 5/12,
a) What is the probability of losing the race?
1 – probability of winning
P(Ā) = 1- P(A) = 1 −
5
12
=
7
12
b) What are odds against winning?
𝑂(Ā) =
P(Ā)
P(A)
=
7
12
5
12
⁄ =
7
5
𝑜𝑟 7: 5
c) If the payoff odd is listed as 6:1, how much profit do you make if you bet $10 and
you win?
Payoff odds against event A = (net profit): (amount bet)
Net Profit = (Payoff odds) × (amount bet)
(10 )( 6) = $60
16. A classic counting problem is to determine the number of different ways that the letters
of "PERSONNEL" can be arranged. Find that number.
Answer: This is a permutation problem with some identical items:
𝑛!
𝑛1!𝑛2!…𝑛 𝑘!
Requirements:
1. There are n items available, and some items are identical to others.
2. We select all of the n items (without replacement).
3. We consider rearrangements of distinct items to be different sequences.
There are 9 letters: n=9 P: 1 E:2 R:1 S:1 O:1 N:2 L:1
Red:
13/49
Red:
14/50
Blue:
36/50
Red:
12/48
Blue:
34/48
Blue:
35/49
Blue:
36/49
Blue:
36/48
Red:
14/49
Red:
13/48
Red:
13/48
Red:
14/48
Blue:
35/48
Blue:
35/48

5. 5
9!
1! 2! 1! 1! 1! 2! 1!
= 90,720
17. When two different people are randomly selected (from those in your class), find the
indicated probability (assume birthdays occur on the same day of the week with equal
frequencies).
a. Probability that two people are born on the same day of the week.
No particular day is specified, the first person can be born on any day.
𝑝(2𝑛𝑑 𝑝𝑒𝑟𝑠𝑜𝑛 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑑𝑎𝑦) =
1
7
𝑝(𝑏𝑜𝑡ℎ 𝑝𝑒𝑟𝑠𝑜𝑛𝑠 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑑𝑎𝑦) =
7
7
(
1
7
) =
1
7
b. Probability that two people are both born on Monday.
𝑝(1𝑠𝑡 𝑝𝑒𝑟𝑠𝑜𝑛 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑀𝑜𝑛) =
1
7
𝑝(2𝑛𝑑 𝑝𝑒𝑟𝑠𝑜𝑛 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑀𝑜𝑛) =
1
7
𝑝(𝑏𝑜𝑡ℎ 𝑝𝑒𝑟𝑠𝑜𝑛𝑠 𝑏𝑜𝑟𝑛 𝑜𝑛 𝑀𝑜𝑛) =
1
7
(
1
7
) =
1
49
18. How many different auto license plates are possible if the plate has:
a) 2 letters followed by 4 numbers?
262
× 104
b) 3 letters – no repeats, followed by 3 numbers - repetition allowed?
26 × 25 × 24 × 103
c) 4 letters – repetition allowed, followed by 2 numbers – no repeats?
264
× 10 × 9
d) 4 places – each character is either a letter or a number?
364
19. In a first grade school class, there are ten girls and eight boys. In how many ways can:

6. 6
a. the students finish first, second and third in a foot race? (assume no ties)
𝑃𝑟
𝑛
= 𝑃3
18
= 18 × 17 × 16
b. the girls finish first and second in a geography contest? (assume no ties)
𝑃𝑟
𝑛
= 𝑃2
10
= 10 × 9
c. three boys be selected for lunch duty?
𝐶𝑟
𝑛
= 𝐶3
8
=
8!
3! (5!)
= 56
d. six students be selected for a hockey team?
𝐶𝑟
𝑛
= 𝐶6
18
=
18!
6! (12!)
= 18,564
e. five students be selected: 3 boys and 2 girls?
𝐶3
8
× 𝐶2
10
=
8!
3! (5!)
∙
10!
2! (8!)
= 2520
f. four girls be selected for a field trip?
𝐶𝑟
𝑛
= 𝐶4
10
=
10!
4! (6!)
= 210