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Chapter 16: Problem 35
35\. Four candidates are running for mayor of Happyville. According to thepolls candidate \(A\) has a "one in five" probability of winning [i.e.,\(\operatorname{Pr}(A)=1 / 5] .\) Of the other three candidates, all we know isthat candidate \(C\) is twice as likely to win as candidate \(B\) and thatcandidate \(D\) is three times as likely to win as candidate \(B\). Find theprobability assignment for this probability space.
Short Answer
Expert verified
The probabilities of each candidate winning are as follows: A: 1/5, B: 2/15, C: 4/15, D: 2/5.
Step by step solution
01
Introduction
Let \(B\), \(C\), and \(D\) represent the probabilities of candidates B, C, and D winning respectively. It's given that \(A\) has a probability of \(1/5\) to win and \(C\) is twice as likely as \(B\) to do so, while \(D\) is three times likely as \(B\). So, we can say that \(C = 2B\) and \(D = 3B\). The sum of all probabilities must equal 1, hence we formulate the equation \(1/5 + B + 2B + 3B = 1\).
02
Solve the equation
To find the values of \(B\), \(C\), and \(D\), solve the equation. Simplify by combining similar terms to form \(6B + 1/5 = 1\). Then, subtract \(1/5\) from both sides of the equation to isolate \(6B\) on one side and get \(6B = 4/5\). Divide both sides by 6 to solve for \(B\). Thus, \(B = (4/5) / 6\), or simplified, \(B = 2/15\).
03
Calculate the probabilities for C and D
Now calculate the probabilities for \(C\) and \(D\) using the values for \(B\) determined in step 2 and the relationships given in the problem. As mentioned above, \(C = 2B\) and \(D = 3B\). Therefore, \(C = 2 * (2/15) = 4/15\), and \(D = 3 * (2/15) = 6/15\), which simplifies to \(D = 2/5\).
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