MATH 110 Module 3 Exam (2025) / MATH110 Module 3 Exam: Portage Learning, Exams of Nursing

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MATH 110 Module 3 Exam (2025) / MATH110 Module 3

Exam: Portage Learning(Questions & Answers)

MATH 110 Module 3 Exam

Exam Page 1

Suppose A and B are two events with probabilities:

P(Ac^ )=.40,P(B)=.45,P(A∪B)=.60.

Find the following: a) P(A∩B).

p(anb) = p(a) + p(b) - p(aub) p(a) = 1 - p(a^c) p(a^c) = 0. 1 - 0.40 = 0. p(a) = 0.

0.60 + 0.45 - 0.60 = 0.

P(AnB) = 0.

b) P(A).

p(a) = 1 - p(a^c) p(a^c) = 0.

1 - 0.40 = 0.

P(A) = 0.

c) P(Bc).

p(b) = 1 - p(b^c) rearranged to find p(b^c) = 1 - p(b)

1 - 0.45 = 0.

Exam Page 3

Find the answer to each of the following by first reducing the fractions as much as possible:

a) P(850,4)=

p(n,r) = n! / (n-r)!

n = 850

r = 4

p(850, 4) = 850! / 846!

850x849x848x847 = 518329442400

Suppose you are going to make a password that consists of 4 characters chosen from {2,7,8,c,f,k,t,z}. How many different passwords can you make if you cannot use any character more than once in each password?

p(n,r) = n! / (n-r)! n = 8 r = 4 8 - 4 = 4

p(8,4) = 8! / 4! 8x7x6x5 = 1680

P(8,4) = 1680 1680 different passwords can be made out of the 8 characters given without one character showing up more than once

Answer Key

Find the answer to each of the following by first reducing the fractions as much as possible: a) P(850,4)=

b) C(530,4)=

P(850,4) = 518329442400

b) C(530,4)=

c(n,r) = n! / r! (n-r)!

n = 530

r = 4

c(530,4) = 530! / (4!)(526!)

(530x529x528x527) / (4x3x2x1) =

C(530,4) = 3250609780

Answer Key

Exam Page 5

In a tri-state conference, 60% attendees are from California, 25% from Oregon, and 15% from Washington. As it turns out 6 % of the attendees from California, 17% of the attendees from Oregon, and 12% of the attendees from Washington came to the conference by train. If an attendee is selected at random and found to have arrived by train, what is the probability that the person is from Washington?

c = california o = oregon w = washington t = train p(c) = 60/100 = 0. p(o) = 25/100 = 0. p(w) = 15/100 = 0. p(t│c) = 0. p(t│o) = 0. p(t│w) = 0.

p(t│w) = ((p(w) x p(t│w)) / ((p(w) x p(t│w)) + ((p(c) x p(t│c)) + ((p(o) x p(t│o)) (0.15 x 0.12) / ((0.15 x 0.12) + (0.60 x 0.06) + (0.25 x 0.17)) 0.018 / (0.018 + 0.036 +0.0425) 0.018 / 0.0965 = 0.

b) What is (B│A)?

Exam Page 6

The probability that a certain type of battery in a smoke alarm will last 2 years or more is .85. The probability that a battery will last 5 years or more is .15. Suppose that the battery is 2 years old and is still working, what is the probability that the battery will last at least 5 years?

e = battery lasting 2 or more years f = battery lasting 5 or more years

p(e) = 0.

In a tri-state conference, 60% attendees are from California, 25% from Oregon, and 15% from Washington. As it turns out 6 % of the attendees from California, 17% of the attendees from Oregon, and 12% of the attendees from Washington came to the conference by train. If an attendee is selected at random and found to have arrived by train, what is the probability that the person is from Washington?

P(Train│C)=.06.. P(Train│O)=.17..

P(Train│W)=.12..

P(C)=.60,P(O)=.25,P(W)=.15.

We want to find P(W│Train), so use:

P(T│W) = 0.

Answer Key

Suppose that 5 out of 11 people are to be chosen to go on a mission trip. In how many ways can these 5 be chosen if the order in which they are chosen is not important.

Since we do not want to count all of the possible orderings, we use combinations.

combo order not important c(n,r) = n! / r!(n-r)! n = r = 5 11 - 5 = 6

c(11,5) = 11! / (5!)(6!) (11x10x9x8x7) / (5x4x3x2x1) 55440 / 120 = 462

C(11,5) = 462 when order is not important, there are 462 ways 5 out of 11 people can be chosen to go on a mission trip.

Answer Key