Asked by sha
if E and F are independent events, find P(F) if P(E)=0.2 and P(E U F)= 0.3
Answers
Answered by
Reiny
If F and E are independent events then P(E and F) = 0
and we also know
P(E or F) = P(E) + P(F) - P(E and F)
.3 = .2 + P(F) - 0
P(F) = .1
and we also know
P(E or F) = P(E) + P(F) - P(E and F)
.3 = .2 + P(F) - 0
P(F) = .1
Answered by
David
Let P(F)= X
since E and F are independent eveents,
P(EnF)= P(E)* P(F)
therefore, P(EnF)= 0.2X
but P(EuF)=P(E)+ P(F)- P(EnF)
0.3= 0.2 + x - 0.2x
x = 1/8
since E and F are independent eveents,
P(EnF)= P(E)* P(F)
therefore, P(EnF)= 0.2X
but P(EuF)=P(E)+ P(F)- P(EnF)
0.3= 0.2 + x - 0.2x
x = 1/8
Answered by
David
sorry, i made a mistake, for mutually exculsive events, P(EnF)= 0, hence,
from the previous equation,
0.3=0.2+P(F)-0
so P(F)= 0.1
from the previous equation,
0.3=0.2+P(F)-0
so P(F)= 0.1
Answered by
Andre
Independent does not mean mutually exclusive, thus
P(EUF)=P(E)+P(F)-P(EnF)
Independence allows us to write:
P(EUF)=P(E)+P(F)-P(E)P(F)
P(EUF)=P(E)+[1-P(E)]P(F)
Therefore,
P(F)= [P(EUF)-P(E)]/[1-P(E)]
P(F)=[0.3-0.2]/[1-0.2]
=.1/.8
=1/8
P(EUF)=P(E)+P(F)-P(EnF)
Independence allows us to write:
P(EUF)=P(E)+P(F)-P(E)P(F)
P(EUF)=P(E)+[1-P(E)]P(F)
Therefore,
P(F)= [P(EUF)-P(E)]/[1-P(E)]
P(F)=[0.3-0.2]/[1-0.2]
=.1/.8
=1/8
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