Question

Consider a probability space where the sample space is Ω = { A,B,C,D,E,F } and the event space is 2 Ω . 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.

Answer #1

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

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