Consider a probability space where the sample space is = { A,B,C,D,E,F } and the
ID: 3068654 • Letter: C
Question
Consider a probability space where the sample space is = { A,B,C,D,E,F } and the event space is 2 . Assume that we only know that the probability measure P {·} satisfies
P ( { A,B,C,D } ) = 4/5
P ( { C,D,E,F } ) = 4/5 .
a) If possible, determine P ( { D } ), or show that such a probability cannot be determined unequivocally.
b) If possible, determine P ( {D,E,F } ), or show that such a probability cannot be determined unequivocally.
c) If possible, determine P ( { E,F } ), or show that such a probability cannot be determined unequivocally.
d) If possible, determine P(A;B;E; F), or show that such a probability cannot be determined unequiv-
ocally.
Explanation / Answer
here as sample space probabiliy is 1.
theefore P(A)+P(B)+P(C)+P(D)+P(E)+P(F) =1 ............(1)
P(A)+P(B)+P(C)+P(D)=4/5 ...............(2)
P(C)+P(D)+P(E)+P(F)=4/5................(3)
a)
as from above variable are 6 and equation are 3 ; therefore we can not solve for P(D).
b)
for abvoe P({D,E,F})=P(D)+P(E)+P(F) ; can be solved from equation (3) ; if we know P(C); but again we can not solve indiviudal P(C) due to equation constraint
c)
adding equation(2) and (3) and then subtracting equation (!) from here:
P(A)+P(B)+P(C)+P(D)+P(C)+P(D)+P(E)+P(F)-(P(A)+P(B)+P(C)+P(D)+P(E)+P(F))=(4/5)+(4/5)-1
P(C)+P(D)=3/5
therefore from equation (3)
P(C)+P(D)+P(E)+P(F) =4/5
3/5+P(E)+P(F) =4/5
P(E)+P(F)=P({E,F})=4/5-3/5=1/5
d)
from equation (2)
P(A)+P(B)+P(C)+P(D) =4/5
P(A)+P(B)+3/5=4/5
P(A)+P(B)=1/5
hence P(A)+P(B)+P(E)+P(F)=P({A,B ,E,F}) =(1/5)+P(1/5)=2/5