Portage Learning MATH 110 Module 3 Exam - VERIFIED 2022

Portage Learning MATH 110 module 3 Exam


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) = 0.40
1 - 0.40 = 0.60
p(a) = 0.60


0.60 + 0.45 - 0.60 = 0.45


P(AnB) = 0.45


b) P(A).


p(a) = 1 - p(a^c)
p(a^c) = 0.40


1 - 0.40 = 0.60


P(A) = 0.60


c) P(Bc).


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


1 - 0.45 = 0.55



This study source was downloaded by 100000841689952 from CourseHero.com on 05 -27-2022 01:54:37 GMT -05:00


https://www.coursehero.com/file/32320826/Module-3-Examdocx/

, P(B^c) = 0.55




Answer Key
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).


For P(A∩B). Use P(A∪B)=P(A)+P(B)-P(A∩B) and rearrange to
P(A∩B)=P(A)+P(B)-P(A∪B). But for this equation, we need P(A) which we can find by using P(A)=1-
P(A^c ). So, P(A)=1-.40= .60.



P(A∩B)=.60+.45-.60=.45.


b) P(A).


P(A) was found above as .60.


c) P(Bc).


For P(Bc ). Use P(B)=1-P(Bc ) which may be rearranged to (Bc )=1-P(B).



P(Bc )=1-.45=.55.


Exam Page 2
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?



This study source was downloaded by 100000841689952 from CourseHero.com on 05 -27-2022 01:54:37 GMT -05:00


https://www.coursehero.com/file/32320826/Module-3-Examdocx/