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

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 } ) = 1/2

P ( { C,D,E,F } ) = 1/2 .

a) If possible, determine P ( { C } ), or show that such a probability cannot be determined unequivocally.

b) If possible, determine P ( { A,B } ), or show that such a probability cannot be determined unequivocally.

c) If possible, determine P ( { B,C } ), or show that such a probability cannot be determined unequivocally.

Explanation / Answer

as we kow that sum of all probabilty in sample space =1

theefore P(A)+P(B)+P(C)+P(D)+P(E)+P(F) =1

here as P({A,B,C}) = P(A)+P(B)+P(C) =1/2

and P({C,D,E,F}) =P(C)+P(D)+P(E)+P(F) =1/2

adding above 2 equations:

P(A)+P(B)+P(C)+P(C)+P(D)+P(E)+P(F) =1/2+1/2

P(A)+P(B)+P(C)+P(D)+P(E)+P(F)+P(C) =1

1+P(C) =1

P(C)=0

b)

P({A,B,C})=P({A,B})+P(C) =1/2

P({A,B})+0 =1/2

P({A,B}) =1/2

c)P({B,C})=P(B)+P(C)=P(B)

here as from abvoe equation we have only equation P({A,B})+0 =1/2; and there are two variables therefore we can not solve for A and B.

hence we can not solve for P ( { B,C } )

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