20. The following two-way contingency table gives the breakdown of the populatio
ID: 3355016 • Letter: 2
Question
20. The following two-way contingency table gives the breakdown of the population in a particular locale according to party affiliation (A, B, C, or None) and opinion on a bond issue: Opinion Favor Oppose Undecid 0.12 0.09 0.07 Affiliati on ed Affiliati on Opinion Favor Oppose Undecid ed 0.16 0.12 0.14 0.04 0.03 0.06 0.08 0.06 0.03 None A person is selected at random. Find the probability of each of the following events. a. The person is affiliated with party B b. The person is affiliated with some party c. The person is in favor of the bond issue. d. The person has no party affliation and is undecided about the bond issue.Explanation / Answer
Ans a. The probability of a random person being affiliated with party B can be found by summing up all the entries in the row of Party B
P(B) = 0.16 + 0.12 + 0.14 = 0.42 (The answer can be found by summing up the entries in the 2nd row)
Ans b. The probability of a person being affiliated to some party is equal to the sum of prpbabilities of a person being affiliated to party A, B and C.
Thus, P(some party) = P(A) + P(B) + P(C)
Now, P(A) = 0.12 + 0.09 + 0.07 = 0.28 (The sum of the entries of the first row)
P(B) = 0.16 + 0.12 + 0.14 = 0.42 (The sum of the entries of the second row)
P(C) = 0.04 + 0.03 + 0.06 = 0.13 (The sum of the entries of the third row)
Thus, P(some party) = 0.28 + 0.42 + 0.13 = 0.83
Ans c. The probability that a random person is in favour of the bond issue can be found by summing up the entries in the first column, ie the column of entries in favour of the bond issue. Thus :
P(favour of bond issue) = 0.12 + 0.16 + 0.04 + 0.08 = 0.40
Ans d. The probability that a random person has no party affiliation and is undecided about the bond issue can be found at the intersection of the "None" row and "Undecided" column.
Thus, P(no party affiliation and undecided about bond issue) = 0.03
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