Stat 151
Test 1
Name___Answer Key__
Please answer all questions carefully. On the first page, circle the desired letter (you may include support work
on attached papers). For all others follow directions and show support work whenever practical.
1. For any two events A and B,
P ( A  B) = P(A) + P(B) - P ( A  B)
2. How many permutations of size 3 can be constructed from the set (A, B, C, D, E)?
5P3
𝟓!
= 𝟐! = 𝟔𝟎
3. If P(B) = 0.40, P ( A  B) = 0.45 and P(A/B) = 0.75, then
P(A/B) =
𝑷 ( A  B)
𝑃(𝐵)
therefore (0.75)(0.4) = 0.3 = P ( A  B)
P ( A  B) = P(A) + P(B) - P ( A  B)
0.45 = P(A) + 0.4 – 0.3 and thus P(A) = 0.35
4. Every hospital has backup generators for critical systems should the electricity go out. A hospital installs n
independent but identical backup generators each of which as a probability of working of p = 0.8. How
large must n be so that the probability that at least one generator operating correctly is no less than 0.9992
P(Generator works) = 0.8 and thus P(fail) = 0.2
If two generators, P(at least one is working) = 1 – P(both fail) = 1 – (0.2)^2
For n generators, P(at least one is working) = 1 – P(all fail) = 1 – (0.2)^n
What is n such that 0.999 < 1 – (0.2)^n
Which is (0.2)^n < 0.001
𝐥𝐧⁡(𝟎.𝟎𝟎𝟏)
And n > 𝐥𝐧⁡(𝟎.𝟐) which means n > 4.29, so there must be at least five generators.
5. Two processes of a company produce rolls of pressed metal: Those produced by process A are flawless 97%
of the time, while process B has a defective rate of only 1%. Three of every five units produced by the
company are using process A with process B being used for the others. If a roll is selected at random and it is
found to be defective, what is the conditional probability that it was produced using process B?
P(F|A) = Probability the roll is flawless given that it was produced using process A = 0.97
P(Fc|A)= Probability the roll is flawed given that it was produced using process A = 0.03
P(F|B) = 0.99
P(Fc|B)= Probability the roll is flawed given that it was produced using process A = 0.03
P(A) = 0.6 and P(Ac) = P(B) = 0.4
𝐂
𝐏(𝐅 |𝐁)𝐏(𝐁)⁡
(𝟎.𝟎𝟏)(𝟎.𝟒)
.𝟎𝟎𝟒
𝟒
c
c
Want P(B|F ), which is given by P(B|F ) = 𝐂
= (𝟎.𝟎𝟏)(𝟎.𝟎𝟒)+(𝟎.𝟎𝟑)(𝟎.𝟔) = .𝟎𝟐𝟐 = ⁡ 𝟐𝟐
𝐂
𝐏(𝐅 |𝐁)𝐏(𝐁)+𝐏(𝐅 |𝐀)𝐏(𝐀)
6. At a popular restaurant, the probability of ordering neither an appetizer nor a dessert is 0.21. For these diners,
the probability of ordering an appetizer is 0.41 and the probability of ordering a desert is 0.53. What is the
probability of ordering both an appetizer and a dessert? Please draw a diagram representing the situation above.
Please also use the following notation and use words to describe what the others mean.
P(A) = orders an appetizer
P(Ac) = does not order an appetizer
P(B) = orders a dessert
P(A∩B) = Orders and appetizer and a dessert
P(Ac∩B) = Does not order an appetizer, but does order a dessert
P(A∩B)c = Does not order both an appetizer and a dessert. = orders at most one of the two = P(Ac U Bc)
7. How many different pizzas can be ordered if there are two sizes, three types of crust and up to eight different
toppings? You can assume that one can order zero to all eight toppings.
(2)(3)(28) = 1536
8. A small grocery store has eight cartons of milk, two of which are sour. If you purchase the fifth carton sold
that day, what is the probability that you get a carton of sour milk?
The probability is always two out of eight or 0.25, no matter when you pick.
9. One lifeboat has three females and two males. A nearby and bigger lifeboat has three males and four females.
One survivor swims from the smaller boat to the bigger one and it is from this boat that the lone survivor is
rescued. What is the probability that the survivor is female?
𝟑
𝟐
𝟒
Method A. Boat 1 has 𝟓 F and 𝟓 M, while boat 2 has 𝟕 F and
𝟑
𝟕
M.
𝟒
Case 1. M swims over (probability of this is 0.4) then P(survivor is F|M) is 𝟖
𝟓
Case 2. F swims over (probability of this is 0.6) then P(survivor is F|F) is 𝟖
𝟐𝟑
Thus total probability is P(F) = P(F|M)P(M) + P(F|F)P(F) = (0.4)(0.5) + (0.6)(0.625) = 𝟒𝟎.
Method B. after a person swims over, the gender distribution is 3.4 male and 4.6 female.
𝟒.𝟔
𝟒.𝟔
𝟐𝟑
Thus probability of female is 𝟒.𝟔+𝟑.𝟒 = ⁡ 𝟖 = ⁡ 𝟒𝟎
10. Paper is often test for “burst strength” and “tear strength.” These strengths are classified as low, medium
and high. After examining a large batch of papers, the following results are tabulated.
Burst strength
Tear Strength
B1 (low)
21
33
36
A1 (low)
A2 (med.)
A3 (high)
B2 (med.)
33
63
27
B3 (high)
39
27
21
If one of the papers is chosen at random from the batch, what are the probabilities that it has the following
characteristics:
𝟐𝟏+𝟑𝟑+𝟑𝟗
a. A1
=
b. A3 ∩ B2
=
c. A1 given B2
=
d. B2 given A3
=
𝟐𝟏+𝟑𝟑+𝟑𝟗+𝟑𝟑+𝟔𝟑+𝟐𝟕+𝟑𝟔+𝟐𝟕+𝟐𝟏
=
𝟗𝟑
𝟑𝟎𝟎
𝟐𝟕
𝟑𝟎𝟎
𝟑𝟑
𝟑𝟑+𝟔𝟑+𝟐𝟕
𝟐𝟕
𝟑𝟔+𝟐𝟕+𝟐𝟏
𝟑𝟑
= 𝟏𝟐𝟑
𝟐𝟕
= 𝟖𝟒
11. Given a set of three eight sided dice, where die 1 has three green sides and five gold sides, die 2 has four of
each color and die three has five green sides and three gold sides, what is the probability that at least one side
comes up gold when they are rolled? What is the probability that exactly one gold side comes up?
𝟑
𝟒
𝟓
𝟒𝟓𝟐
P(at least one gold) = 1 – P(no Gold) = 1 – (𝟖) (𝟖) (𝟖) = (𝟓𝟏𝟐)
P(exactly one gold) = P(just first is gold) + P(just second is gold) + P(just third is gold)
𝟓
𝟒
𝟓
𝟑
𝟒
𝟓
𝟑
𝟒
𝟑
𝟏𝟗𝟔
= (𝟖) (𝟖) (𝟖) + (𝟖) (𝟖) (𝟖) + (𝟖) (𝟖) (𝟖) = (𝟓𝟏𝟐)
12. For events A & B, the probability of event A not happening is 0.2 and event B is equally likely to happen or
not. If the probability of either A or B occurring is 0.9, are A and B independent events?
P(Ac) = 0.2 thus P(A) = 0.8. Further, P(B) = 0.5. P(A or B) = P(A U B) = 0.9
To show independence, we need to show that P(A and B) = P(A)P(B).
Since P(A and B) = P(A) + P(B) – P(A or B), we just need to show that P(A) + P(B) – P(A or B) =
P(A)P(B).
And, (0.8) + (0.5) – (0.9) = 0.4 which equals, (0.8)(0.5), so it is true.
13. An urn contains n balls numbered from 1 through n. A sample of size n is selected from the urn one at a
time. A match occurs if the ball numbered i is selected on the ith draw.
a. For n = 1 to 5, find the probability of at least one match if the sampling is done (i) with replacement and (ii)
without replacement.
b. how much does n affect the probabilities?
c. How does sampling with and without replacement affect these probabilities?
If n = 1
If n = 2
If n = 3
If n = 4
If n = 5
P(match w/o rep.) = 1 out of 1
P(match w/o rep.) = 1 out of 2
P(match w/o rep.) = 4 out of 6
P(match w/o rep.) = 16 out of 24
P(match w/o rep.) = out of 120
&
&
&
&
&
P(match with rep.) = 1 out of 1.
P(match with rep.) = 3 out of 4.
P(match with rep.) = 19 out of 27.
P(match with rep.) = 175 out of 256.
P(match with rep.) = 2101 out of 3125.
for n
P(match w/o rep.) = ? out of n!
& P(match with rep.) = (nn – (n-1)n) out of nn.