Ask Question
20 January, 08:41

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 fraction of voters participated in the local election?

+4
Answers (1)
  1. 20 January, 09:25
    0
    a) 0.161

    b) 0.186

    c) 0.1392

    d) 2431/5000

    Step-by-step explanation:

    Voters

    Independents - 46% =

    Liberals - 30% =

    Conservatives - 24% =

    Elections:

    Independents - 35% = 0.35

    Liberals - 62% = 0.62

    Conservatives - 58% = 0.58

    what is the probability that he or she is

    (a) an Independent?

    0.46*0.35 = 0.161

    (b) a Liberal?

    0.3*0.62 = 0.186

    c) a Conservative?

    0.24*0.58 = 0.1392

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

    0.161 + 0.186 + 0.1392 = 0.4862 = 4862/10000 = 2431/5000
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A total of 46 percent of the voters in a certain city classify themselves as Independents, whereas 30 percent classify themselves as ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers