MATH 110 LATEST 2023 Graded A+

MATH 110 LATEST 2023 Graded A+

Chapter 3 math 110
Exam Page 1
Suppose A and B are two events with probabilities:



P(A)=.35,P(Bc )=.45,P(A∩B)=.25.


Find the following:
a) P(A∪B).


P (AUB)=P(AP+P (B)-P(A⋂B)
P(B∁) => P(B)=1-P(b∁)=1-0.45=0.55
P(AUB)=0.35+0.55-0.25=0.65


b) P(Ac ).


P(A∁)=> P(A)=1-P(A∁ )=>
P(A∁)=1-P(A)=1-0.35=0.65

c) P(B).




P(B∁)=> P(B)=1-P(B∁)
P(B)=1-0.45=0.55



Instructor Comments
Very good.
Answer Key
Suppose A and B are two events with probabilities:



P(A)=.35,P(Bc )=.45,P(A∩B)=.25.


Find the following:
a) P(A∪B).
This study source was downloaded by 100000812546443 from CourseHero.com on 07-25-2023 02:34:11 GMT -05:00


https://www.coursehero.com/file/67176820/Chapter-3-math-110docx/

, MATH 110 LATEST 2023 Graded A+


For P(A∪B). Use P(A∪B)=P(A)+P(B)-P(A∩B) . But for this equation, we need P(B) which we can find
by using P(B)=1-P(Bc ). So, P(B)=1-.45= .55.


P(A∪B)=.35+.55-.25=.65

b) P(Ac ).


For P(Ac ). Use P(A)=1-P(Ac ) which may be rearranged to (Ac )=1-P(A) .


P(Ac )=1-.35=.65.


c) P(B).




For P(B). Use (B)=1-P(Bc ) .



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


Exam Page 2
Suppose you are going to make a password that consists of 5 characters chosen from
{2,4,9,b,d,g,k,m,s,w}. How many different passwords can you make if you cannot use any character
more than once in each password?


P(10,5)=10!/(10-5)=10!/5=10(9)(8)(7)(6) =30,240 potential password combinations



Instructor Comments
Very good.
Answer Key
Suppose you are going to make a password that consists of 5 characters chosen from
{2,4,9,b,d,g,k,m,s,w}. 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 100000812546443 from CourseHero.com on 07-25-2023 02:34:11 GMT -05:00


https://www.coursehero.com/file/67176820/Chapter-3-math-110docx/