Probability Problem

prettylittlepixels

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It is election day on McCoy Isle. 8% of the residents live in rural areas (R), 13% live in Littleville (L), 27% live in Middletown (M), and 52% live in Hugecity (H). There are three candidates running for the office of Overlord: Spendthrift Sue (S), Balanced-Budget Bob (B), and Environmental Ed (E).

Of the rural residents (GIVEN RURAL), 20% favor Sue, 15% favor Bob, and 65% favor Ed.
Of the Littleville residents, 20% favor Sue, 25% favor Bob, and 55% favor Ed.
Of the Middletown residents, 30% favor Sue, 25% favor Bob, and 45% favor Ed.
Of the Hugecity residents, 40% favor Sue, 45% favor Bob, and 15% favor Ed.

You can answer these questions using just the probability formulas, but most students will find a contingency table or a tree diagram helps to organize your thoughts.
I made the following contingency table

Sue (S)
Bob (B)
Ed (E)
Rural (R) .08
.2
.15
.65
Littleville (L) .13
.2
.25
.55
Middletown (M) .27
.3
.25
.45
Hugecity (H) .52
.4
.45
.15
Total Probabilities
.331
.346
.323

If every resident votes and the candidate with the most votes becomes Overlord, who will be the new Overlord?
-Sue
-There will be a tie.
-Ed
-Bob

Bob because .346 is the largest probability on my contingency table:
(.08*.15)+ (.13*.25)+ (.27*.25)+ (.52*.45)= 0.346
Sue’s came out to .331 and Ed’s came out to .323

If an individual is chosen at random, what is the probability that the person is a Littleville resident and favors Bob?

.13*.25= 0.0325

If a resident is chosen at random, what is the probability that the person is either a resident of Hugecity or favors Sue or both?

I’m pretty sure I didn’t do this problem right. I think the word or in the problem is throwing me off because I calculated:
P(H and S)= .52 * .331= 0.17212
P(H or S)= P(H) + P(S) – P(H and S)= .52 + .331 - .17212= .67888

So I then added .67888+ .17212= 0.851 but this doesn’t sound right. The question says is either a resident of Hugecity or favors Sue or both. The 2 or’s in this problem confuse me. Does it mean I should add P(H) + P(S) + P(H and S)= .52+.331+.17212= 1.02312, but this can’t be right because the probability can’t be more than 1…

A randomly-chosen resident favors Ed. What is the probability that the person is a rural resident?

For this problem I took P(E) * P(R)= .323 * .08= .02584 or should I do .08/.323= .2477?

A randomly-chosen resident favors Bob. What is the probability that the person is a resident of Middletown?

Same as the part of the question above. P(B) * P(M)= .346*.27= .09342 or is this not right?

A randomly-chosen citizen favors Sue. What is the probability that the citizen is NOT a resident of Littleville?

For this problem I:
1-.13= .87
.87*.331= .28797?
 
It is election day on McCoy Isle. 8% of the residents live in rural areas (R), 13% live in Littleville (L), 27% live in Middletown (M), and 52% live in Hugecity (H). There are three candidates running for the office of Overlord: Spendthrift Sue (S), Balanced-Budget Bob (B), and Environmental Ed (E).

Of the rural residents (GIVEN RURAL), 20% favor Sue, 15% favor Bob, and 65% favor Ed.
Of the Littleville residents, 20% favor Sue, 25% favor Bob, and 55% favor Ed.
Of the Middletown residents, 30% favor Sue, 25% favor Bob, and 45% favor Ed.
Of the Hugecity residents, 40% favor Sue, 45% favor Bob, and 15% favor Ed.

You can answer these questions using just the probability formulas, but most students will find a contingency table or a tree diagram helps to organize your thoughts.
I made the following contingency table
Sue (S)Bob (B)Ed (E)
Rural (R) .08.2.15.65
Littleville (L) .13.2.25.55
Middletown (M) .27.3.25.45
Hugecity (H) .52.4.45.15
Total Probabilities.331.346.323

If every resident votes and the candidate with the most votes becomes Overlord, who will be the new Overlord?
-Sue
-There will be a tie.
-Ed
-Bob

Bob because .346 is the largest probability on my contingency table:
(.08*.15)+ (.13*.25)+ (.27*.25)+ (.52*.45)= 0.346
Sue’s came out to .331 and Ed’s came out to .323

If an individual is chosen at random, what is the probability that the person is a Littleville resident and favors Bob?

.13*.25= 0.0325

If a resident is chosen at random, what is the probability that the person is either a resident of Hugecity or favors Sue or both?

I’m pretty sure I didn’t do this problem right. I think the word or in the problem is throwing me off because I calculated:
P(H and S)= .52 * .331= 0.17212
P(H or S)= P(H) + P(S) – P(H and S)= .52 + .331 - .17212= .67888 This is the answer

So I then added .67888+ .17212= 0.851 but this doesn’t sound right. The question says is either a resident of Hugecity or favors Sue or both. The 2 or’s in this problem confuse me. Does it mean I should add P(H) + P(S) + P(H and S)= .52+.331+.17212= 1.02312, but this can’t be right because the probability can’t be more than 1…

A randomly-chosen resident favors Ed. What is the probability that the person is a rural resident?

For this problem I took P(E) * P(R)= .323 * .08= .02584 or should I do .08/.323= .2477? I didn't get either one of these choices.

A randomly-chosen resident favors Bob. What is the probability that the person is a resident of Middletown?

Same as the part of the question above. P(B) * P(M)= .346*.27= .09342 or is this not right? Not right, I think

A randomly-chosen citizen favors Sue. What is the probability that the citizen is NOT a resident of Littleville?

For this problem I:
1-.13= .87
.87*.331= .28797?
Consider P(H or S), in terms of a Venn diagram. One circle is Sue, with area = 0.313, and the other circle is H, with area 0.52. The intersection of the two circles is P(H and S) = 0.17212.

If you add together the two circles, you will have counted the intersection twice. so you have to subtract it once:
P(H or S) = P(H) + P(S) – P(H and S)= .52 + .331 - .17212 = .67888
That is the answer .. you have counted H and S, including "both of the above" once

Of the .323 who favor Ed, 0.08*.65 (or 0.052) are Rural. P(R | E) = P(R and E) / P(E)

Likewise P(M | B) = P(M and B) / P(B)

You might have done the last three better if you used a different form of the ocntingency table. If you normalize the rows so they add up to the population represented by that row, it would look like this.

Sue (S)Bob (B)Ed (E)
Rural (R) .08.016
.012
.052
Littleville (L) .13.026
.0325
.0715
Middletown (M) .27.081
.0675
.1215
Hugecity (H) .52.208
.234
.078
Total Probabilities.331.346.323

Now every entry is a fraction of the total population. The columns add up to the vote fraction, and the rows add up to the resident fraction. Does that make the conditional probabilities easier to see? If you add up all of Ed's votes, you get .323. Of the four boxes that contribute to Ed, the fraction if the R box is .052/.323 = 0.161.

Sue's column adds up to .331, of which .026 is from L, so .305 is notL. P(notL | S ) = P(notL and S) / P(S)
 
for (2)

P[H and S] + P[H or S] = P(H) + P(S) = 0.52 + .331 = 0.851

You can convince yourself of this by picking some population of this area and putting numbers to your percentages.

If there are a total of one million people. Then 520K live in Huge City. 331K are pro Sue. If a person is both a resident of Hugecity and favors Sue it will be one of these people. So (520K + 331K) = 851K satisfy the conditions which corresponds to the Pr=0.851 as found above.

Thanks for that tip... I had never thought to substitute a population in as you did above, but that explains it much better and is easier to understand! The Statistics software I have for my class that contains these practice questions for my final next week is not accepting my answer of .8510 for this question though? It won't tell me the correct answer, but just shows it's wrong.


for (3)

Use Bayes Law.

Pr[A | B] Pr = Pr[B | A] Pr[A]

Pr[resident favors Ed | resident is rural] Pr[resident is rural] = Pr[resident is rural | resident favors Ed] Pr[resdient favors Ed]

Pr[resident is rural | resident favors Ed] = Pr[resident favors Ed | resident is rural] Pr[resident is rural]/Pr[resident favors Ed]

(.65 * .08) / .323 = .161

(4) is almost identical to (3) so give it a shot again.

(.25*.27) / .346= .1951

(5) find the probability she is using what we've done above, and then subtract this from 1 to get the probability she isn't.

(.2*.13) / .331= .0785498489

1- .0785498489= .9215


so redo 3-5 by using Bayes Law
 
If a resident is chosen at random, what is the probability that the person is either a resident of Hugecity or favors Sue or both?

I’m pretty sure I didn’t do this problem right. I think the word or in the problem is throwing me off because I calculated:
P(H and S)= .52 * .331= 0.17212
P(H or S)= P(H) + P(S) – P(H and S)= .52 + .331 - .17212= .67888 This is the answer


I submitted my study guide with all of my work to my professor to prepare for my final next week and she said this part is wrong? She won't give the answer, but just says if it is right or wrong for me to fix. I am at a loss here... Help!
 
Looking at the Venn diagram I've convinced myself the answer to this is

Pr[person is from Hugecity] + Pr[person favors Sue] - Pr[person is from Hugecity AND favors Sue] =

.52 + .331 - .4 = 0.451

I apologize for my earlier error.

No need to apologize. I appreciate the help! :)
 
you should try to reproduce the relevant part of the venn diagram for this with the associated probabilities and convince yourself it's correct. It's not hard.

I did and I that really helped because it showed me that I needed to:

Take the probability of Hugecity or (H), which was .52 and add it to the total probabilities I added for Sue or (S), which was .331 and then I needed to subtract .4 or the 40% of (H) residents that favor Sue.

This gave me .52+.331-.4= .451

THANKS!
 
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