Show via probability manipulation that the following is true: P(A∩B∩C) = P(A|B∩C) x P(B|C) x P(C)
Solution:
Conditional probability rule shows that
P(A|B)= P(A∩B)/P(B)
P(A ∩ B) = P(A | B)P(B) = P(B | A)P(A)......multiplication rule.
So, P(A∩(B∩C))= P(A|(B∩C))* P(B∩C)..........(i)
In the same manner by the multiplication rule:
P(B∩C) = P(B|C) x P(C).....(ii)
Putting the (ii) in to (i), we get
P(A∩B∩C) = P(A|B∩C) x P(B|C) x P(C)
Show via probability manipulation that the following is true: P(A∩B∩C) = P(A|B∩C) x P(B|C) x P(C)
1 pt) A P(X1126) Probability B. P(X < 966) Probability c. P(X > 1046) Probability
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