# Conditional probability if 40\% of the population have completed college, and 85\% of college graduates are registered to vote, what percent of the population areboth college graduates andregistered voters? To find: The percent of people who is both college graduates and registered voters.

Question
Upper level algebra
Conditional probability if $$\displaystyle{40}\%$$ of the population have completed college, and $$\displaystyle{85}\%$$ of college graduates are registered to vote, what percent of the population areboth college graduates andregistered voters?
To find: The percent of people who is both college graduates and registered voters.

2021-02-23
Approach:
Conditional probability is the probability of an event occurring based on the occurrence of a previous event.
In other words, $$\displaystyle{P}{\left({B}{\mid}{A}\right)}$$ is a conditional probability that the event B will occur given that another event A has already occurred. In this case, A and B are dependent events.
$$\displaystyle{P}{\left({B}{\mid}{A}\right)}$$ is ead as the conditional probability of B given A.
Multiplication property of probabilities: For events A and B, $$\displaystyle{P}{\left({A}\cap{B}\right)}={P}{\left({A}\right)}\cdot{P}{\left({B}{\mid}{A}\right)}={P}{\left({B}\right)}\cdot{P}{\left({A}{\mid}{B}\right)}$$.
Given:
$$\displaystyle{40}\%$$ of the population is college graduates and $$\displaystyle{85}\%$$ of college graduates are registered voters.
Calculation:
Let P(A) be the probability that a person have completed college and let $$\displaystyle{P}{\left({B}{\mid}{A}\right)}$$ is the probability that a college graduates are registered to vote.
According to question, $$\displaystyle{P}{\left({A}\right)}={0.4}$$ and $$\displaystyle{P}{\left({B}{\mid}{A}\right)}={0.85}$$.
The probability that a person is both college graduates and registered voters is $$\displaystyle{P}{\left({A}\cap{B}\right)}$$.
The probability of occurrence of both A and Bcan be found by using multiplication property of probabilities.
Substitute $$\displaystyle{P}{\left({A}\right)}={0.4}$$ and $$\displaystyle{P}{\left({B}{\mid}{A}\right)}={0.85}$$ in $$\displaystyle{P}{\left({A}\cap{B}\right)}={P}{\left({A}\right)}-{P}{\left({B}{\mid}{A}\right)}$$.
$$\displaystyle{P}{\left({A}\cap{B}\right)}={\left({0.4}\right)}\cdot{\left({0.85}\right)}={0.34}$$
Final statement:
Therefore, the population is both college graduates and registered voters is $$\displaystyle{34}\%$$.

### Relevant Questions

The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
The presidential election is coming. Five survey companies (A, B, C, D, and E) are doing survey to forecast whether or not the Republican candidate will win the election. Each company randomly selects a sample size between 1000 and 1500 people. All of these five companies interview people over the phone during Tuesday and Wednesday. The interviewee will be asked if he or she is 18 years old or above and U.S. citizen who are registered to vote. If yes, the interviewee will be further asked: will you vote for the Republican candidate? On Thursday morning, these five companies announce their survey sample and results at the same time on the newspapers. The results show that a% (from A), b% (from B), c% (from C), d% (from D), and e% (from E) will support the Republican candidate. The margin of error is plus/minus 3% for all results. Suppose that $$\displaystyle{c}{>}{a}{>}{d}{>}{e}{>}{b}$$. When you see these results from the newspapers, can you exactly identify which result(s) is (are) not reliable and not accurate? That is, can you identify which estimation interval(s) does (do) not include the true population proportion? If you can, explain why you can, if no, explain why you cannot and what information you need to identify. Discuss and explain your reasons. You must provide your statistical analysis and reasons.
To calculate: The probability that a student chosen randomly from the class is not going to college and on the honor roll if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76, out of these 56 are going to college.
To calculate: The probability that a student chosen randomly from the class is not going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76,out of these 56 are going to college.
To calculate: The probability that a student chosen randomly from the class is going to college if the number of students in a high school graduating class is 128, out of which 52 are on the honor roll, and out of these, 48 are going to college. The number of students who are not on the honor roll is 76, out of these 56 are going to college.

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
1. A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 44.7 texts per day. The standard deviation was 16.5 texts. a. To compute the confidence interval use a ? z t distribution. b. With 98% confidence the population mean number of texts per day is between and texts. c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population number of texts per day and about percent will not contain the true population mean number of texts per day. 2. You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately $$\displaystyle\sigma={40.4}$$ dollars. You would like to be 90% confident that your estimate is within 1.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? n = 3. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately $$\displaystyle\sigma={57.5}$$. You would like to be 95% confident that your estimate is within 0.1 of the true population mean. How large of a sample size is required?
Solid NaBr is slowly added to a solution that is 0.010 M inCu+ and 0.010 M in Ag+. (a) Which compoundwill begin to precipitate first? (b) Calculate [Ag+] when CuBr justbegins to precipitate. (c) What percent of Ag+ remains in solutionat this point?
a) AgBr: $$\displaystyle{\left({0.010}+{s}\right)}{s}={4.2}\cdot{10}^{{-{8}}}$$ $$\displaystyle{s}={4.2}\cdot{10}^{{-{9}}}{M}{B}{r}$$ needed form PPT
CuBr: $$\displaystyle{\left({0.010}+{s}\right)}{s}={7.7}\cdot{\left({0.010}+{s}\right)}{s}={7.7}\cdot{10}^{{-{13}}}$$ Ag+=$$\displaystyle{1.8}\cdot{10}^{{-{7}}}$$
b) $$\displaystyle{4.2}\cdot{10}^{{-{6}}}{\left[{A}{g}+\right]}={7.7}\cdot{10}^{{-{13}}}$$ [Ag+]$$\displaystyle={1.8}\cdot{10}^{{-{7}}}$$
c) $$\displaystyle{\frac{{{1.8}\cdot{10}^{{-{7}}}}}{{{0.010}{M}}}}\cdot{100}\%={0.18}\%$$
A multiple regression equation to predict a student's score in College Algebra $$\displaystyle{\left(\hat{{{y}}}\right)}$$ based on their high school GPA $$\displaystyle{\left({x}_{{{1}}}\right)}$$, their high school Algebra II grade $$\displaystyle{\left({x}_{{{2}}}\right)}$$, and their placement test score $$\displaystyle{\left({x}_{{{3}}}\right)}$$ is given by the equation below.
$$\displaystyle\hat{{{y}}}=-{9}+{5}{x}_{{{1}}}+{6}{x}_{{{2}}}+{0.3}{x}_{{{3}}}$$
A multiple regression equation to predict a student's score in College Algebra $$(\hat{y})$$ based on their high school GPA (x1x1), their high school Algebra II grade (x2x2), and their placement test score (x3x3) is given by the equation below.
$$\hat{y}=-9+5x1x1+6x2x2+0.3x3x3$$