Probability with random?

Question

In a survey of 1000 eligible voters selected at random, it was found that 62% had a college degree.

Additionally, the survey results showed tha of the eligible voters who had a college degree, 85% voted in the most recent senatorial election.

Furthermore, only 42% of the eligible voters without a college degree voted in the same election.

Based on this information and assuming it is representative of the general voting population, find the probability that a
randomly selected eligible voter

a. voted in the last senatorial election.

b. has a college degree and did not vote in the last senatorial election.

c. does not have a college degree and did not vote in the last senatorial election.

How can I solve this? :-|
Thanks.

Answer 1

a. Separate the eligible voters into those who have a college degree and those who do not by the probability of each. Since 62% of them have a degree, 38% do not.

Degree: 62/100
No Degree: 38/100

Now, calculate the probabilities of a person having a degree and voting in the election, and not having a degree and voting in the election separately.

Degree: (62/100) * (85/100) = 5270/10000
No Degree: (38/100) * (42/100) = 1596/10000

Sum these two values, and you have the probability of a person being in either group and voting in the election.

Voting: (5270/10000) + (1596/10000) = 6866/10000 = 3433/5000. As a percentage, 68.66%.


b. We already know that 62% of eligible voters have a college degree. Now we need to calculate the probability of having a college degree and not voting in the election. This is done in a similar manner as step two of part a.

Since 85% of people with a college degree voted in the last election, it is obvious that 15% of those people did not.

Degree and No Vote: (62/100) * (15/100) = 930/10000 = 93/1000. As a percentage, 9.3%.


c. This is done just like part b, but using the group of voters without degrees. Since 42% of them voted, 58% did not.

No Degree, No Vote: (38/100) * (58/100) = 2204/10000 = 551/2500. As a percentage, 22.04%.

Answer 2

take event C as voter had a college degree,
and event E as he/she participate in recent election.
^ : intersection
U : union
| : conditional
' : compliment / negation

P(C) = 0.62
P(C | E) = 0.85
P(E^C') = 0.42 ---> P(E ^ C) = 1 - 0.42 = 0.58

a. Probability that a randomly voter vote,
P(E) = ...?
*
Bayes theorem:
P(C | E) = P( E ^ C) / P(E)
P(E) = P(E ^ C) / P(C | E) = 0.58 / 0.85
= 0.682352941
Expected Value of E :
Mean(E) = 1000 * 0.682 = 682
.: 682 eligible voters voted in last election

b. has college degree and did note vote:
P(C^E') = ...?
**
Assume events C and E are uncorrelated and so independent
P(C^E) = P(C) * P(E) = 0.62 * (1-0.682352941)
= 0.196941177
Expected value of C^E:
Mean (C^E') = 0.197 * 1000
= 197
.: 197 college graduates did not vote

c. doesn't have college degree nor vote:
P(C'^E') = ...?

P(C'^E') = P(C') * P(E') = (1-0.62) * (1-0.682352941)
= 0.120705882
Expected vale\ue of C'^E':
Mean (C'^E') = 0.121 * 1000 = 121
.: 121 alligibles who did not vot don't have any college degree

You can learn more from

Probability & Statistics for Engineers & Scientists written by Ronald E. Walpole.
Probability and Statistics written by Athanasios Papoulis

Answer 3

a. 68.66%
b. 9.3%
c. 22.04%

calculated
a.
(62% x 85%) = 52.7% (percentage that have degree and voted)
38% x 42% = 15.96% (percentage that have no college degree and voted)

52.7%+15.96% = 68.66%

b.
62% (have college degree) x 15% ( percentage not voting) = 9.3%

c.
38% ( no college degree) x 58% ( percent not voting) = 22.04%

Answer 4

Let
P(c) = probability of having college degree
P(v) = probability of voting in last election

Given

P(c) = 0.62
P(v|c) = 0.85
P(v|~c) = 0.42

a) Calculate P(v).

P(~c) = 1 - P(c) = 1 - 0.62 = 0.38

P(v) = P(v|c)P(c) + P(v|~c)P(~c) = 0.85*0.62 + 0.42*0.38
P(v) = 0.6866
________________

b) Calculate P(c∩~v).

P(~v|c) = 1 - P(v|c) = 1 - 0.85 = 0.15

P(c∩~v) = P(~v|c)P(c) = 0.15*0.62 = 0.093
_______________

c) Calculate P(~c∩~v).

P(~v|~c) = 1 - P(v|~c) = 1 - 0.42 = 0.58

P(~c∩~v) = P(~v|~c)P(~c) = 0.58*0.38 = 0.2204