AP Stats Unit 4 Practice FRQ #3

A large group of voters in two nearby counties were surveyed about which political party they prefer between parties X, Y, and Z. The relative frequencies in the table below summarize the results of the survey.

a. If one surveyed person is selected at random:
i. What is the probability that the person lives in County A and prefers Party Z?
ii. What is the probability that the person lives in County A or prefers Party Z?
iii. What is the probability that the person lives in County A, given that they prefer Party Z?

b. In the survey, are the events “Lives in County B” and “Prefers Party X” independent? Explain why or why not.

c. Assume that the distribution of party preference shown in the table holds true in the entire state. If 5 voters from the state are selected at random, what is the probability that exactly 2 of them prefer Party Y?

a) i. P(County A and Party Z) = 0.152
There is a 0.152 probability that the person lives in County A and prefers Party Z.

ii) P(County A or Party Z) = P(County A) + P(Party Z) - P(County A and Party Z)
P(County A or Party Z) = 0.4 + 0.31 - 0.152 = 0.558
There is a 0.558 probability that the person lives in County A or prefers Party Z.

iii) P(County A|Party Z) = P(County A and Party Z) / P(Party Z) = 0.152 / 0.31 = 0.49
There is a 0.49 probability that the person lives in County A, given that they prefer Party Z.

b) If independent, P(Lives in County B and Prefers Party X) = P(Lives in County B) * P(Prefers Party X).

0.108 = 0.6 * 0.18 = 0.108

The events “Lives in County B” and “Prefers Party X” are independent, as the two events satisfy the multiplication rule for independent events.

c) Binomial Distribution Conditions - Binary - Party Y or not Party Y, Independent - reasonable to assume preferences of 5 voters are independent from each other, Number of Trials = 5 voters are selected at random (n=5), S = probability of selecting a voter who prefers Party Y is the same for every voter selected (p = 0.51)

X = the number of voters who prefer Party Y when the distribution of X is binomial with p = 0.51 and n = 5

P(X=2) = 0.306 using binomialpdf with p = 0.51, n = 5, and successes = 2.

There is a 0.306 probability that exactly two voters prefer Party Y if 5 voters from the state are selected at random.

Hi Brandon -

Well done! All three parts show strong command of statistical notation, and clear communication. Your answers are all correct and would all earn credit.

~Jerry

a.)
i. P(County A AND Party Z)= 0.152
There is a 15.2% chance that the person lives in County A and prefers Party Z.

ii.) P(County A OR Party Z)=0.400+0.310-0.152=0.558
There is a 55.8% chance that the person lives in County A or prefers Party Z

iii.) P(County A|Party Z)=0.152/0.310=0.490
There is a 49% chance that the person lives in County A given that they prefer Party Z.

b.) P(County B)= 0.600, P(County B|Party X)=0.108/0.180=0.60
Because the two probabilities are the same, the events “Live in County B” and “Prefers Party X” are independent.

c.) This scenario has each trial ends is either a success or failure, they are independent, the 2 selected are set, and the probability for those who prefer Party Y is the same at 0.510. Therefore, this is a binomial probability.
P(X=2)= BinomPDF(n=5, p=0.510, x=2)=0.306
There is a 30.6% chance that if 5 voters from the state are selected at random, exactly 2 of them will prefer Party Y.

Welcome back!

In all three parts, your answers are correct and presented with accompanying work and appropriate notation. You would get full credit for each part! Part ( c ) provides a good contrast to your work on Unit 4 FRQ #2 - by saying “the probability for those who prefer Party Y is the same at 0.510,” you provide an in-context reason for the scenario being binomial. Well done!

A.
i. P(County A and Party Z)=.152/1=.152
ii. P(County A or Party Z)= .400+.310-.152=.558
iii. P(County A/Party Z)= .152/.310=.490

B. Yes, “lives in county B” and “prefers party x” are independent because:
P(County B/Party X)= P(County B): P(.6)= P(.600)
P(Party X/County B)= P(Party X): P(.18)= P(.180)

C. The probability that exactly 2 of them prefer Party Y is .306.
To find out I first check the binomial distribution requirements:

  1. Each trial has 2 possible outcomes (Party Y or not party Y)
  2. There are fixed number of trials (n=5)
  3. Outcomes are independent
  4. Probability of success is the same for each trial (p=.510)

I then proceeded to put on my calculator binompdf( 5, .510, 2). I knew to use the function binompdf because the problem is asking to find exactly 2 voters.

Once again, nice work! All calculations and explanations are correct, with appropriate work shown. However, in part (b), you make a small notation error. P(County B | Party X) = P(County B) = 0.6. The probabilities themselves should not be inside the P( ) notation. Think of it like input/output function notation - inside the P( ) is what probability you’re finding (in words or symbols), while outside on the other side of the equals sign is the numerical answer.

~Jerry

a i) The probability that a person lives in Country A and prefers Party Z is 0.152.
ii) The probability that the person lives in County A or prefers Party Z is 0.558.
P(AUB) = P(person lives in county A) + P(Person prefers Party Z) - P(person who lives in Country A and prefers Party Z)= 0.558
iii) P(Person lives in County A | Person prefers Party Z) = 0.152/0.310= 0.490

b) P(Living in County B)= 0.600 P(Living in County B | Prefers Party X)=0.108/ 0.180= 0.6
The events “Lives in County B” and "Prefers Party X"are independent as the probability of Living in County B equals the probability of living in county B given preferring Party X.

c. Binomial Distribution conditions:
Binary: Prefer Party Y or not prefer Party Y.
Independent: Sample size of 5 voters is less that 10% of the population(voters from the entire state)
Number of trials: There are a set number of trials (5) n=5
Same probability of success: p= 0.510

Calculator: P(x=2)= Binom (n=5, p=0.50, x=2) = 0.306
If 5 voters from the state are selected at random, the probability that exactly 2 of them prefer Party Y is 0.306.

Nice job once again! Small thing: for part a-ii, you provide the correct answer and use correct notation. However, you don’t show the numbers that go into the calculation, so on some rubrics that would be considered a “naked” answer in that you didn’t show the numbers along the way. Taking the extra step to show calculations with probability scenarios is always helpful. Everything else should receive full credit.

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a)
i. P(county A and prefers party z)=.152 --> There’s a 15.2% chance that a person lives in county A and prefers party z.

ii. P(county A or prefers party z)= P(county A) + P(party z) - P(county A and party z) --> .4 + .31 - .152=.558 or 55.8% --> There’s a 55.8% chance that a person lives in county A or prefers party Z.

iii. P(county A|party z)= P(county A and party z)/P(party z) --> .152/.31=.49 --> There’s a 49% chance that the person lives in county A given that they prefer party Z.

b) P(Lives in county B|party x) --> P(county b and party x)/P(party x) = P(county B) <-- if this is true, the two events are independent

P(.108)/P(.18)=.6 which equals P(.6)
This statement is true, which shows that these two events are independent of each other.

c) This is a binomial random variable because there is 2 outcomes for each trial – prefer party y (.51) or don’t prefer party y (.49). There is a set number of trials – 5 randomly selected voters from this state. The trials are independent of each other – seen in part b. Lastly, the probability of a success will not change in each trial (p=.51).

P(exactly 2/5 of randomly selected voters from the state prefer party y) = (5 2) = 5!/2!(5-2)! --> 10(.51)^2(.49)^3 = .306 --> There’s a 30.6% chance that exactly 2/5 of the randomly selected voters from the state prefer party Y.

Thank you so much for your feedback on my answers! I really appreciate it.

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a) i) the probability that a person lives incountry A and prefers party Z is 0.152
ii) the probability that a person lives in country A or prefers party Z is 0.4+0.31-0.152 = 0.558
iii) the probability that they live in country A given that they prefer party Z is 0.152/0.31 = 0.49032
b) for the events of living in country B and preferring party X to be independent the proabbility that a person lives in country B and the probability. that. they live in B given that. they prefer party X has to be equal. P(B) = 0.6 P(BㅣX) = 0.6549
since the probabilities are not equal, the two events are not independent.
c) binomial probability
binompdf (trials : 5, p : 0.51, x value : 2) = 0.306

a)
—i)
P(AZ)=0.152
—ii)
P(AUZ)=P(A)+P(Z)-P(A∩Z)=0.400+0.310-0.152=0.558
—iii)
P(AIZ)=P(A∩Z)/P(Z)=0.152/0.310=0.490323
b)
When the situation is Independent P(A∩B)=P(A)×P(B)
0.108=0.600×0.180
So the events “Lives in County B” and “Prefers Party X” are Independent.
c)
We know this is a Binomial Distribution since it’s Binary, Independent, & fixed number of trials.
binomCdf(n=5,p=0.510,x=2)=0.306005

Nice job on your calculations (and super shout-out for going by hand - or at least writing it all out by hand - on the binomial calculations). Be really careful with your notation on part (b). Probability notation is kind of like function notation f(x) from algebra. The “stuff” inside the parentheses is like an input - and should represent what we’re finding the probability of. This will include variables and equality/inequality statements in some cases, but not just “plain numbers”, since those are the “output” in this situation. So P(.108) isn’t a thing - it should be P(B and X) = .108. That is, you already wrote the notations correctly - just put the actual numbers on your second line for part (b), and you’ll be good to go to say yes for independence.

Hello again!

On part (a) and part ( c ), you’ve done all calculations correctly. For part (b), you’ve defined what we need to find to show independence… but your calculation P(B | X) is incorrect. You should get 0.108 / 0.18 = 0.6, so it would be independent. Double check what you did. Given what probability you got, you made an appropriate conclusion, so you’d still be eligible for partial credit.

Good on all three parts!

a.
i. P(AandZ)=.152
ii. P(AorZ)=.400+.310-.152=.558
iii. P(AgivenZ)=.152/.31=.49
b. Yes they are independent because P(BandX)=.108 and P(B)*P(X)=.108. Since these two numbers are the same then the events are independent.
c. This is a binomial distribution because there is a success or failure as to whether the voters prefer Party Y or not, each voter chosen is independent of one another, Number of trials is fixed at 5, and there is a set number of probabilities for each choice. Binompdf(trials=5,p=.510,x=2)=.306
There is a .306 probability that exactly 2 of the 5 voters prefer Party Y.

Nice job again!

In part (b), you should do a little more to show where the P(B)*P(X) = 0.108 comes from. P(B and X) comes directly off the table so there’s no additional work to show, but P(B)*P(X) required a calculation, so you should show the numbers you multiplied. All other parts have appropriate work and corect answers.

a.
i. There is a 15.2% chance that the person lives in County A and prefers party Z
ii. There is a 55.8% chance that the person lives in County A or prefers Party Z.
iii. There is about a 49% chance that a person lives in County A, given that they prefer Party Z.

b. Yes, the events “Lives in County B” and “Prefers Party X” are independent. For two events to be independent, the P(A) must be equal to the P(A|B). The P(Lives in County B) is 60%, and then the P( Lives in County B | Prefers Party X)= .108/.18=.6=60%

c. P(X=2) = ( 5 2 )(.51)^2(1-.51)^5-2= .306

If 5 voters from the state are selected at random, the probability that exactly 2 of them prefer Party Y is 30.6%

a.
i.P(A and Z) = .152
There is a .152 probability that a randomly selected person lives in Country A and prefers Party Z.
ii. P(A or Z)= p(a) +p(z)- overlap or p(a and z) so .4+.310-.152 = .558
There is a .558 probability that a randomly selected person lives in Country A or prefers Party Z.
iii. P(a|z)= p(a and z) / P(z) or .152/.310 = .490
There is a .490 probability that a randomly selected person lives in Country A given that they prefer Party Z.
b. Independence means the probability of one variable is not affected if we know the other or in context P(lives in country B) = P(lives in country b|prefers party x)
p(b)= .6
p(b|x) = .108
.6 does not equal .108 so the events lives in country b and prefer party x are not independent.
c. P(2/5 prefer y)
n= 5 randomly selected people
p = .510 chance of success of preferring party y
x= 2 people exactly given to prefer party y
Each trial independent as same probability of success each time, fixed number of trials(5), and outcome is success or failure(person doesn’t prefer party y)
binompdf(5,.51,2)= .36
There is a .36 probability that exactly 2/5 voters selected at random prefer Party Y

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