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/
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/
