Practice Voters In A Town

harpazo

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Mar 20, 2018
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There are 1265 eligible voters in a town, and 972 of them are registered to vote. If one eligible voter is selected at random, what is the probability that this voter is a. registered b. not registered? Do these two probabilities add up to 1.0? If yes, why?
 

Kayculator

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Jan 30, 2020
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I'm probably tackling this completely wrong but feel free to correct me completely.
If one eligible voter is selected at random, what is the probability that this voter is registered?
76.84% or 0.7684

what is the probability that this voter is not registered?
23.16% or 0.2316

Do these two probabilities add up to 1.0? If yes, why?
Yes. They add to 1.0 because we are only drawing one voter and the two conditions are mutually exclusive. Each voter either is registered or is not.


JUSTIFICATION:
We say 1265 eligible voters is 100% of them. The easiest way for me to see the percentage of registered voters, is to put is as a fraction. Remembering that the 100% means 1.0 because \(\displaystyle \left(\frac {1265}{1265}\right) = 1\).

We know that the registered voters are a slice of the total. 972 out of 1265, is the fraction the voters that are registered. In other words, \(\displaystyle \left(\frac {972}{1265}\right)\) of the total voters are registered. And passing that fraction as a decimal we get a 0.7684 percentage the selected voter to be registered.

Getting the amount of not registred voters is pretty easy now that we have the percentage since we're parting from the assumption both should add to that 1.0 percentage. So the total minus the registered voters should give us the amount of no registered voters.

\(\displaystyle 1.0000 - 0.7684 =\) 0.2316 percentage that the selected voter will not be registered.

If you want to do it the way as the first answer. Then we can make the fraction. We know that whatever of the 1265 voters minus the 972 registred voters are not registered, so
\(\displaystyle \left(\frac {1265-972}{1265}\right)\) which means 293 out of 1265, or \(\displaystyle \left(\frac {293}{1265}\right)\). Again passing to decimal we'll get 0.2316. Same result.


I'm sure there is some fraction looking formula I don't have with me to calculate probabilities, but I thought I'd give it a go anyway.
 
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MarkFL

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I agree with your results completely. :)