Question

events E, F, and G in a sample space S. Assume that Pr[E]=0.4, Pr[F]=0.45, Pr[G]=0.45, Pr[E∪F]=0.6, Pr[E∪G]=0.65, and Pr[F∪G]=0.7. Find the following probabilities:
(1) Pr[E′∪F]=


(2) Pr[F′∩G]=


(3) Pr[E∩G]=

Answer 1

okay

Answer 2

P(E∪G)=p(E)+P(G)-P(E∩G)
0.65=0.4+0.45-P(E∩G)
P(E∩G)=0.85-0.65=0.2

Answer 3

is that for (F'∩G)

Answer 4

now we make make a list of probabilities:
Pr[E]=0.4,
Pr[F]=0.45,
Pr[G]=0.45,
Pr[E∪F]=0.6,
Pr[E∪G]=0.65,
Pr[F∪G]=0.7.
P(E∩G)=0.2
P(F∩G)=0.2
p(E∩G)=0.2

Answer 5

(1) Pr[E′∪F]= .6?


(2) Pr[F′∩G]= .2?


(3) Pr[E∩G]= .2

Answer 6

@amirreza1870 ?

Answer 7

now we start to find P(E′∪F)=P(E′)+P(F)-P(E'∩G)
THE third one is true but we are asking for P(E'∩G) not p(E∩G)

Answer 8

ok

Answer 9

sorry.P(E′∪F)=P(E′)+P(F)-P(E'∩F)
.P(E′∪F)=0.6+0.45-P(E')*P(F)
.P(E′∪F)=0.6+0.45-0.27=0.78
i used these rules:P(A)=1-P(A')
when A and B are independent :P(A∩B)=P(A)*P(B)

Answer 10

how would you find #2 then?

Answer 11

P(F′∩G)=P(F')*P(G)

Answer 12

did you make a venn diagram?

Answer 13

.78 is wrong

Answer 14

when two events are independent P(A∩B)=P(A)*P(B)

Answer 15

(1) Pr[E′∪F]= 0.78
(2) Pr[F′∩G]= 0.2475
(3) Pr[E∩G]= 0.2

Answer 16

1 isnt right

Answer 17

every human can make mistakes.can you explain why the first one isn't correct?

Answer 18

is it because of P(E')*P(F)

Answer 19

but it's Pr[E′∪F]

Answer 20

where did .27 come from?

Answer 21

p(E'∩F)=P(E')*P(F)=(1-P(E))*P(F)=(1-0.4)*0.45=0.6*0.45=0.27

Answer 22

oh

Answer 23

but then I don't see why its the wrong answer

Answer 24

for the first one,
\(P[E\cup F] = P[F]+P[E]-P[E\cap F]\)
\( P[E\cup F] -P[F]= P[E]-P[E\cap F]\)
\(P[E' \cap F] = 1-(P[E]-P[E\cap F]) =1- P[E\cup F] +P[F]\)

Please draw a venn diagram if you cannot clarify