Question

Proofs

a) With conditional probability, P(A|B), the axioms of probability hold for the event on the left side of the bar. A useful consequence is applying the complement rule to conditional probability. We have that P(A|B) = 1 − P(A|B). Prove this by showing that P(A|B) + P(A|B) = 1 (Hint: just use the definition of conditional probability)

b) If two events A and B are independent, then we know P(A ∩ B) = P(A)P(B). A fact is that if A and B are independent, they so are all combinations of A, B, . . . etc. Show that if events A and B are independent, then P(A ∩ B) = P(A)P(B), and thus A and B are independent. (Hint: P(A ∩ B) = 1 − P(A ∪ B). Then use addition rule and simplify.)

Answer 1

@ Jo prove PCAB= 1-PCAB), we proceed as below - Multiplication (definition Applying the orem i of conditional probabili we have PTAIB) = PCAMB) PCB Now, PCAUĀTB)=P(AVĀ NB) using 12 - PCB) A -PAUA PCB) T' AUA and PC) B, are independent - PCAUAJ As we know if A and B Now, we get are independent then PCAVA/B) = PTAUĂ) PCAMB) =P CA)PCB). PIAVA | B] =P(A/B) IPTABL e Tucing a de Also, AUA=5, is decotes the entire sample space. .PCAUA) = P(S) = T theorem of I probability for disjoin events e.ePTAUB = PIA +PIB HIMANSHI

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