Question

If A is a subset of B and B is a subset of C and:
P(A|B) = 0.6 and P(B|C) = 0.6 and P(C) = 0.6, then find:
P(AUB), P(not B|C), P(B|not A)

Answer 1

I been trying to this for a while, when i made the sets, i made 3 circles inside each other, and said if P(C) = 0.6 then P(B) must be 0.36 since P(B|C) = 0.6 and P(A) = 0.216 since P(A|B) = 0.6, am i going about this the wrong way, please help, thank you

Answer 2

You have that right.

Answer 3

Hmm okay so P(AUB) = P(A) + P(B) - P(AB) in this case since A is a susbet of B then P(AB) = P(A) since P(AUB) = P(B) = 0.36. But I am still stuck on the other two

Answer 4

Given \[A \subset B \subset C\]
And P(C)=0.6 then P(B)=P(B intersects C)=0.36 and P(A)=P(A intersects B)=0.216

Therefore\[P(A \cup B)=P(A)+P(B)-P(A intersects B)=P(A)+P(B)-P(A)=0.36\]
[P(B'|C)
=P(B' intersects C)/P(C)
=[P(C)-P(B)]/P(C)
=1-P(B)/P(C)
=1-0.36/0.6=2/5\]

P(B|A')
=P(B intesects A')/P(A')
=[P(B)-P(A)]/P(C)-P(A)
=(0.36-0.216)/(0.6-0.216)
=0.375