MATH 110 Module 3 Exam Questions and Answers – Portage Learning.

MATH 110 Module 3 Exam Questions and Answers – Portage Learning. Exam Page 1 Suppose A and B are two events with probabilities: P(Ac )=.40,P(B)=.45,P(A∪B)=.60. Find the following: a) P(A∩B). p(anb) = p(a) + p(b) - p(aub) p(a) = 1 - p(a^c) p(a^c) = 0.40 1 - 0.40 = 0.60 p(a) = 0.60 0.60 + 0.45 - 0.60 = 0.45 P(AnB) = 0.45 b) P(A). p(a) = 1 - p(a^c) p(a^c) = 0.40 1 - 0.40 = 0.60 P(A) = 0.60 c) P(Bc ). p(b) = 1 - p(b^c) rearranged to find p(b^c) = 1 - p(b) 1 - 0.45 = 0.55 P(B^c) = 0.55 Answer Key Suppose A and B are two events with probabilities: P(Ac )=.40,P(B)=.45,P(A∪B)=.60. Find the following: a) P(A∩B). For P(A∩B). Use P(A∪B)=P(A)+P(B)-P(A∩B) and rearrange to P(A∩B)=P(A)+P(B)-P(A∪B). But for this equation, we need P(A) which we can find by using P(A)=1- P(A^c ). So, P(A)=1-.40= .60. P(A∩B)=.60+.45-.60=.45. b) P(A). P(A) was found above as .60. c) P(Bc ). For P(Bc ). Use P(B)=1-P(Bc ) which may be rearranged to (Bc )=1-P(B). P(Bc )=1-.45=.55.