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Statistics / A First Course in Probability 9 / Chapter 3 / Problem 18P

A First Course in Probability | 9th Edition | ISBN: 9780321794772 | Authors: Sheldon Ross

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A total of 46 percent of the voters in a certain city

Chapter 3, Problem 18P

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QUESTION:

Problem 18P

A total of 46 percent of the voters in a certain city classify themselves as Independents, whereas 30 percent classify themselves as Liberals and 24 percent say that they are Conservatives. In a recent local election, 35 percent of the Independents, 62 percent of the Liberals, and 58 percent of the Conservatives voted. A voter is chosen at random. Given that this person voted in the local election, what is the probability that he or she is

(a) an Independent?

(b) a Liberal?

(c) a Conservative?

(d) What percent of voters participated in the local election?

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QUESTION:

Problem 18P

A total of 46 percent of the voters in a certain city classify themselves as Independents, whereas 30 percent classify themselves as Liberals and 24 percent say that they are Conservatives. In a recent local election, 35 percent of the Independents, 62 percent of the Liberals, and 58 percent of the Conservatives voted. A voter is chosen at random. Given that this person voted in the local election, what is the probability that he or she is

(a) an Independent?

(b) a Liberal?

(c) a Conservative?

(d) What percent of voters participated in the local election?

ANSWER:

Step 1 of 4

Given a total number of percent of the voters is 46.

Our goal is

a). We need to find an Independent.

b). We need to find a Liberal.

c). We need to find a Conservative.

d). We need to find percent of voters participated in the local election.

Let I, L and C be the event that a random person is an independent, liberal or a conservative respectively.

Given P(I) = 0.46, P(L) = 0.3 P(c) = 0.24 and

P $\left({\frac{V}{I}}\right)$ = 0.35, P $\left({\frac{V}{L}}\right)$ = 0.62, P $\left({\frac{V}{C}}\right)$ = 0.58

Now we have to compute P $\left({\frac{I}{V}}\right)$ , P $\left({\frac{L}{V}}\right)$ , and P $\left({\frac{C}{V}}\right)$ .

We know that from Bayes’ rule,

P $\left({\frac{I}{V}}\right)$ = $\frac{P\left({\frac{V}{I}}\right)\ P(I)}{P(V)}$

P $\left({\frac{I}{V}}\right)$ = $\frac{P\left({\frac{V}{I}}\right)\ P(I)}{P\left({\frac{V}{I}}\right)\ P(I)+P\left({\frac{V}{L}}\right)\ P(L)+P\left({\frac{V}{C}}\right)\ P(C)}$

But, P(V) = $P\left({\frac{V}{I}}\right)\ P(I)+P\left({\frac{V}{L}}\right)\ P(L)+P\left({\frac{V}{C}}\right)\ P(C)$

Then P(V) is

P(V) = $(0.35)\times{}(0.46)+(0.62)\times{}(0.3)+(0.58)\times{}(0.24)$

P(V) = 0.161+0.186+0.1392

P(V) = 0.4862

Therefore, the probability of voters is 0.4862.

a). Now we have to calculate an independent.

$P\left({\frac{I}{V}}\right)=$ $\frac{(0.35)\times{}(0.46)}{0.4862}$

$P\left({\frac{I}{V}}\right)=$ $\frac{0.161}{0.4862}$

$P\left({\frac{I}{V}}\right)=$ 0.3311

Therefore, $P\left({\frac{I}{V}}\right)$ is 0.3311

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