Use the set information provided below to answer questions 1-5 Universal set 1 2 3 4 5 6 7 8 9 10 11 12 Set A 1 2 3 4 5 6 7 8 9 10 Set B 1 3 4 7 8 9 Set C 1 4 6 8 9 11 12 Set D 1 4 6 7 8 9 10 11 12 Find the elements and the probability of: 1 P(A?C) 2 P(AUC)' 3 P(A/C) 4 P(BUD) 5 P[(AnB)UC]

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Step 1/5 Given universal set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} and number of element in universal set is n(S) = 12 and subset is SA= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} SB = {1, 3, 4, 7,8,9} So= {1, 4, 6, 8, 9, 11, 12} SD= {1, 4, 6, 7, 8,9, 10, 11, 12} Explanation To find the probability from the given set and subset is very easy.Step 2/5 A 1). Intersection of subset A and C is given by AnC = {1, 4, 6,8,9} i.e. number of element in intersection term is "(An C) = 5 So the probability is P(AnC) = m(Anc) n(S) 5 12 = 0.417 Explanation Here we use classical definition of probability.Step 3/5 2). Union of set A and C is AUC = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} m(A UC) = 12 This union is equal to universal set, Therefore P(A UC) = (A UC) n(S) 12 12 =1 and P(AUC) =1-1 =0 Explanation Probability of complement term is 0.Step 4/5 A 3). Conditional probability is given by P(A|C) P(AnC) P(C) We obtained above P(An C), now probability of P(C) is Since n(C) = 7 P(C) = I(C) n(S) 7 12 Therefore P(A|C) = = 0.714Step 5/5 A 4). Given B UD is obtained from the given set as BUD = {1, 3, 4, 6, 7, 8, 9, 10, 11, 12} 1 (BUD) = (BUD) n(S) 10 12 = 0.833 5. (AnB) = {1,3, 4, 7, 8,9} and T [(AnB) UC] = {1, 3, 4, 7, 8, 9} U {1, 4, 6, 8, 9, 11, 12} IN = {1, 3, 4, 6, 7, 8, 9, 11, 12} IN and number of element is "[(AnB) UC] = 9 Therefore probability is P[(AnB) ug) = (AnB) UC] n(S) 9 12 = 0.75Final answer Therefore probability is P(AnC) = 0.417 P(AUC)' = 0 P(A|C) = 0.714 P(BUD) = 0.83 P[(AnB) UC] = 0.75