[DOCUMENT TITLE]
, Suppose A and B are two events with probabilities:
P(Ac )=.70,P(B)=.65,P(A∪B)=.80.
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(Ac ). So, P(A)=1-.70= .30.
P(A∩B)=.30+.65-.80=.15.
b) P(A).
P(A) was found above as .30.
c) P(Bc).
For P(Bc ). Use P(B)=1-P(Bc ) which may be rearranged to (Bc )=1-P(B) .
P(Bc )=1-.65=.35.
Suppose you are going to make a password that consists of 6 characters chosen from
{3,4,5,7,d,f,g,k,q,r,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 100000817936687 from CourseHero.com on 02 -21-2022 14:09:05 GMT -06:00
https://www.coursehero.com/file/39276415/test-3/
, Suppose A and B are two events with probabilities:
P(Ac )=.70,P(B)=.65,P(A∪B)=.80.
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(Ac ). So, P(A)=1-.70= .30.
P(A∩B)=.30+.65-.80=.15.
b) P(A).
P(A) was found above as .30.
c) P(Bc).
For P(Bc ). Use P(B)=1-P(Bc ) which may be rearranged to (Bc )=1-P(B) .
P(Bc )=1-.65=.35.
Suppose you are going to make a password that consists of 6 characters chosen from
{3,4,5,7,d,f,g,k,q,r,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 100000817936687 from CourseHero.com on 02 -21-2022 14:09:05 GMT -06:00
https://www.coursehero.com/file/39276415/test-3/
