Probability Midterm Quiz ?s

Let X and Y be Bernoulli Random Variables with parameter p, and let Z = X+Y. Is Z also a Bernoulli Random Variable with parameter p?

I think Z is not Bernoulli because it can have more than two outcomes. For example, if X and Y takes on {0,1} with parameter p, Z would take on {0,1,2}.

Consider rolling a fair 6-sided die twice. Let X be the result of the first roll, and let Y be the sum of the two rolls. Which of the following are true? (A) X is a discrete uniform random variable but Y is not. (B) Both X and Y are discrete uniform random variables. (C) Neither X nor Y are discrete uniform random variables. (D) Y is a discrete uniform random variable but X is not.

(A) is correct. X is a discrete uniform random variable because each of the results of the first roll is equally likely since the 6-sided die is fair. However, Y is not a discrete uniform random variable because the sums of the two rolls are not equally likely. For example, the probability of rolling a sum of 12 is not equal to rolling the sum of a 6 because you can only get 12 by rolling 6 twice, but you can get a sum of 6 by rolling (4,2), (2,4), (5,1), (1,5), or (3,3). Thus the outcomes of Y are not all equally likely and Y is not a discrete uniform random variable.

Let E, F, and G be three events. Find an expression for the event so that, of E, F, and G, exactly two of the events occur.

(E ∩ F ∩ Gc) (union) (E ∩ Fc ∩ G) (union) (Ec ∩ F ∩ G)

As in the lecture, consider a system that consists of three parallel units, each of which can be 'up' or 'down' independently of the others. Unit 1 is 'up' with probability p_1 = .6, unit 2 is 'up' with probability p_2 = .4, and unit 3 is 'up' with probability p_3 = .3. What is the probability that there is a path through the system from left to right that is 'up'?

1- (1-P1)(1-P2)(1-P3)

Suppose that we toss two fair dice. Let A be the event that the first die rolls a 4, let B be the event that the sum of the dice is 6 and let C be the event that the sum of the dice is 7. Which of the following is true: - Events A and B are not independent, but events A and C are. - Events A and B are independent, but events A and C are not. - Events A and B are independent, and so are events A and C. - Neither events A and B, nor events A and C are independent.

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You lost your keys... You are 80% certain that the missing keys are in one of the two pockets of the jacket you left in your friend's dorm room, with each pocket being equally likely (that is you are 40% certain they are in the left-hand pocket and 40% certain they are in the right-hand pocket). You friend checks the left-hand pocket and your keys are not there. Given this information, what is the conditional probability that the keys are in the other pocket?

???

Let X be the outcome of the roll of a fair 6-sided die. Find E(X^2) and (E(X))^2.

Again we start by finding the PMF of all of the possible rolls of the 6-sided die. Since the die is fair, every outcome is equally likely so the PMF is 1/6 for every possibility. pX(1)=1/6; pX(2)=1/6; ... pX(6)=1/6 To find (E(X))2, we start by just finding E(X). E(X) = (1/6)(1+2+3+4+5+6) = 21/6 Then, (E(X))2 = (21/6)2 = 12.25 To find E(X2), we square all of our values before we add them together. E(X2)= (1/6)(12+22+32+42+52+62) = 91/6 =15.167

Two coins are flipped. The first coin will land on heads with probability .6, the second with probability .7. Assume that the results of the flips are independent, and let X equal the total number of heads that result from the two flips. Find the expectation of X.

E(X) = 0.6+0.7 = 1.3

Let X be a random variable with expectation E(X) = 10 and variance var(X) = 2. Let Y = 3X-1. What are E(Y) and var(Y)?

E(Y) = 3(E(X))-1 E(Y)= 30-1=29 var(Y)= a^2var(X) = 9var(X)= 18

Two fair dice are rolled (again...). This time, let A be the event that the sum of the dice is 7, and let B be the event that the first die does not equal 4. Which of the following are true: - Events A and B are independent. - Events A and B are not independent. - There is not enough information to tell.

Events A and B are independent. Let's take (a,b) where a is the roll of the first die and b is the roll of the second die. We know that there are 6/36 cases where the sum of the dice is 7. They are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). Of these 6 events, there are 5 instances where the first die does not equal 4. Thus, A intersect B = 5/36. We also know that P(first die does not equal 4)=5/6. 5/6*1/6=5/36. Thus, P(A intersect B) = P(A)P(B). So the two events are independent.

Two six-sided dice each have had two of their sides painted red, two painted black, one painted yellow, and the other painted white. When this pair of dice is rolled, what is the probability that both dice land with the same color face up?

If you draw a matrix with the possible results of the first roll on the x-axis and the possible results of the second roll as the y-axis, then you will see that there will be 10 out of 36 squares shaded in where each roll lands on the same color. Below is the graph, with x's shown for the sequence of rolls fulfilling the solution. r x x r x x b x x b x x y x w x r r b b y w Since there are only 10 possible scenarios out of 36 total scenarios, the probability that both dice will land with the same color face up will be 10/36, or 5/18.

A math test has two questions. You can answer the first question correctly with probability .7, and the second question with probability .6. Question 1 is worth 20 points, and question 2 is worth 30 points. What is the expected number of points you receive on the test? Hint: Let X be the number of points received on the test. Start by finding p_X(0), p_X(20), p_X(30) and p_X(50).

Let A be the event of getting question 1 correct. Let B be the event of getting question 2 correct. pX(0) = P(Ac and Bc) = (.3)(.4) = .12 pX(20) = P(A and Bc) = (.7)(.4) = .28 px(30) = P(Ac and B) = (.3)(.6) = .18 px(50) = P(A and B)= (.7)(.6) = .42 E(X)= (0)(.12)+(20)(5.6)+(30)(5.4)+(50)(.42) = 32

You are rolling a fair, six-sided die and are told that the outcome, X, is even. What is the conditional variance of X, given this information?

Let A be the event that the outcome is even. We want to find var(X|A). So first we must find E[X|A]. Since we have a fair, six-sided die, we know that E[X|A] = 1/3(2+4+6)=4 and E[(X|A)2] = 1/3(4+16+36)=56/3 var(X|A) = E[(X|A)2]-E[X|A]2 = (56/3)-42 = 8/3 = 2.67

Two fair dice are rolled and you are told that they landed on two different numbers. What is the conditional probability that at least one of them landed on a 6?

Let A= Event one landed on a 6 Let B= Event of 2 different #s Looking for P(A I B) = P(A & B)/ (P(B)) In finding P(A & B), we see 5 out of the 36 elements of the sample space (1,6) (2,6) (3,6) (4,6) (5,6) satisfy the desired outcome, so P(A & B) = 5/36 In finding P(B), we can subtract the 6 events in which B does not occur from the 36 total events, to find the number of events in which B does occur. The 6 events B does not occur are (1,1),(2,2),(3,3),(4,4)(5,5)(6,6). Therefore P(B)= 30/36 and so P(A I B) = (5/36)/(30/36) = (1/6)

An urn contains 8 red balls and 4 white balls. You draw two balls from the urn without replacement. At each draw, each ball in the urn is equally likely to be chosen. What is the probability that both balls drawn are red?

Let R1 and R2 be the events that the first and second ball are red, respectively. Then we want P (R1 and R2). P (R1) = 8/12. Also, P (R2|R1) = 7/11 due to the reduced sample space. P(R1&R2)=P(R1)*P(R2|R1)=(8/12)*(7/11)=14/33

You are throwing a fair six-sided die. If the outcome is odd, you win twice the value that appears on the die, and if the outcome is even you win half of the value that appears on the die. What are your expected winnings?

Let X be the winnings, A be the event that the outcome is odd, with X=2Y (Y be the value of the outcome in A). and B be the event that the outcome is even, with with X=Z/2 (Z be the value of the outcome in B). Since an outcome cannot be both odd and even, but must be one of them, we know that A and B are disjoint and their union is all of the sample space. Thus, by the total expectation theorem, we know that E[X]=P(A)·E[X|A] + P(B)·E[X|B] E[X|A] = E[2Y|A] = 1/3(2+6+10) = 6 E[X|B] = E[Z\2|B] = 1/3(1+2+3) = 2 P(A)=P(B)=1/2 So we have E[X]=P(A)·E[X|A] + P(B)·E[X|B] =1/2·6+1/2·2 = 4.

Suppose that a certain disease can be found (randomly) in one percent of the general population, and a blood test has been developed that is 98% accurate in detecting the disease (that is, if a person has the disease and the blood test is performed, then 98% of the time the test will show a positive result). But the test also has false positives in 5% of the cases where the disease is not present. What is the probability that a person does not have the disease and a positive test result?

P(no disease)=.99 P(false positive)=.05. P(no disease and false positive)=.99*.05= .0495

Suppose you roll two dice and know that each of the 36 possible outcomes is equally likely. You observe that the first die is a 3. Given this information, what is the probability that the sum of the two dice equals 8?

P（A）=probability of having a three＝ 1/6 2,6 3,5 4,4 5,3 6,2 P（B）=probability of having an eight = 5/36 P(A&B)=1/36 （3，5） P(B|A)=P(A&B)/P(A)=1/6

Suppose that A and B are mutually exclusive events for which P(A) = 0.3 and P(B) = 0.5. What is the probability that A occurs but B does not?

Since A and B are mutually exclusive, we know that if one occurs then the other cannot occur at the same time. Therefore, the probability that A occurs but B does not is simply what we are given: P(A) = 0.3.

Thirty-one students each flip a fair coin. What is the probability that there are more heads than tails showing?

Since there are an odd number of flips, there is no way for the number of heads to be equal to the number of tails. The coin is fair so there is an equal chance for heads and tails. Therefore the probability of getting more heads than tails is 1/2.

Remember the previous question (the one with the car insurance company)? Here is the setup again: 30% of drivers are prone to accidents, whereas the remaining 70% of drivers are not. Statistically, a driver who is prone to accidents will have an accident within a 1-year period with probability 0.4, whereas a driver who is not prone to accidents has only a probability of 0.2 to get into an accident with a 1-year period. A new policy holder has an accident within a year of purchasing a policy. What is the probability that he/she is accident prone?

So the answer to this question will be a conditional probability: the probability that he/she is accident prone (A) given that he/she had an accident (B). Therefore, we need to calculate P(A|B). We actually calculated P(B) in the previous question since it is the probability that he/she will have an accident. So now we have to find P(A intersection B). But P(A intersection B) will just be the probability that he/she is accident prone and he/she has an accident, which is just 1a from the branch above. Given that, we know that P(A intersection B) = (0.3 * 0.4). P(A|B)=P(A intersection B)/P(B) = (0.3 * 0.4) / (0.3 * 0.4 + 0.7 * 0.2) = 6/13 Therefore, 6/13 is the solution.

Back to our two rolls of a fair, six-sided die. Let again X be the sum of the two rolls. What is P_X(6)?

The best way to solve this problem is to make a 6x6 grid showing the summation of all possible two rolls. Then you add up the number of cells that add up to 6, which is 5.

A power utility can supply electricity to a city from n different power plants. Power plant i fails with probability p_i, independent of the others. Suppose that any one plant can produce enough electricity to supply the entire city. What is the probability that the city will experience a black-out?

The only way the city can black-out is if all n power plants fail, since only one is needed to supply the whole city with electricity. Because their failures are independent of each other, we can just multiply the individual probabilities of each plant failing to get the probability that every plant will fail. Thus, P(black-out)=p1p2p3....pn.

You are setting the table for dinner. The rectangular place mats on your table have dimensions 10 inches x 15 inches, and your round plates have a radius of 3 inches. You make sure that the center of each plate is on a place mat. What is the probability that the entire plate is on the place mat (and does not stick out over the edge of the mat)?

The probability of the entire plate being on the placemat (assuming an equally likely placement at any point on the placemat) is the ratio of the sum of all possible points which satisfy the condition of the whole plate being over the placemat (the area) divided by the area of the placemat. Since the plates have radius 3, the closest they can come to any edge is 3 inches away. Therefore, the center of the plates must be in the centermost 4 inches x 9 inches. Our probability is given by the ratio of these areas, or (4x9)/(10x15)= 6/25

In the example of tossing a biased coin that was just discussed (where heads occurs with probability p and tails with probability 1-p), what is the probability that in five consecutive tosses you obtain a total of four heads?

There are 5 choose 4 ways to order 4 heads out of 5 tosses. Since coin tosses are independent, we can multiply this combination by the probability of getting 4 heads (p^4) by the probability of getting 1 tails (1-p). Thus we have: P(4 heads out of 5 tosses) = (5 choose 4)(p^4)(1-p)

A system is composed of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector (x_1, x_2, x_3, x_4, x_5), where x_i is equal to 1 if component i is working and is equal to 0 if component i is failed. How many outcomes are in the sample space of this experiment?

There are two possible outcomes for each of the five components: 0 or 1. We can list all possible combinations of such outcomes: 0, 0, 0, 0, 0 0, 0, 0, 0, 1 0, 0, 0, 1, 0 0, 0, 0, 1, 1 0, 0, 1, 0, 0 0, 0, 1, 0, 1 ... Altogether we observe that there are 2*2*2*2*2 = 32 possible combinations.

According to a car insurance company, 30% of drivers are prone to accidents, whereas the remaining 70% of drivers are not. Statistically (according to the insurance company) a driver who is prone to accidents will have an accident within a 1-year period with probability 0.4, whereas a driver who is not prone to accidents has only a probability of 0.2 to get into an accident within a 1-year period. What is the probability that a new policy holder will have an accident within a year of purchasing a policy?

Therefore, to get the total probability, we multiply down the branches of each of the possible scenarios. For the accident-prone driver branch, we take (0.3 * 0.4). For the non-accident-prone driver branch, we take (0.7 * 0.2). Adding the probabilities of these two scenarios, we get 0.26 as the total probability that a new policy holder will have an accident within a year of purchasing a policy.

Consider a biased coin with P(Heads) = 0.3. You are tossing this coin 10 times. What is the probability of observing a total of 3 heads in those 10 throws?

This is just a binomial probability. Using the equation (n choose k)pk(1-p)n-k we get: P(3 heads in 10 tosses) = (10 choose 3)(.3)3(.7)7 = 0.267

An urn contains 10 white and 5 black balls. Balls are randomly selected, one at a time, until a black one is drawn. If we assume that each ball selected is replaced before the next one is drawn, what is the probability that exactly n draws are needed?

Use geometric distribution function. We want white balls to come before a black ball. The probability of white balls appearing each time is 10/15 (since ball selected is replaced). Given that we select the balls n times, we are choosing white balls n-1 times before we get a black ball, and it can be written as (10/15)^(n-1). Now, we want the last ball to be a black one and the probability of us getting a black ball in a draw is 5/15. Therefore the probability that exactly n draws are needed is (10/15)^(n-1) * (5/15).

Let X be the outcome of the roll of a fair die. What is var(X)?

Var(X) = E(X^2) - [E(X)]^2 E(X^2) = (1/6) [1+4+9+16+25+36] = 91/6 [E(X)]^2 = ((1/6) [1+2+3+4+5+6])^2 =(21/6)^2 = 49/4 Var(X) = (91/6) - (49/4) = (35/12)

In the example from the lecture of choosing with equal probability one of two unfair coins A and B and then tossing this unfair coin twice (where coin A lands on heads with probability 0.9 and coin B lands on heads with probability 0.1), what is the probability that the second toss is heads, given that the first toss is heads?

We can write this out by using the definition of conditional probability where AH1=first toss of coin A is heads and AH2=second toss of coin A is heads and BH1=first toss of coin B is heads and BH2=second toss of coin B is heads. So we want P((AH2+BH2)|(AH1+BH1))=P((AH2+BH2)∩(AH1+BH1))/P(AH1+BH1) but since they are independent we can say =(P(AH2)P(AH1))+(P(BH2)P(BH1))/(P(AH1)+P(BH1)). So we have (.9*.9)+(.1*.1)/(.9+.1)= 0.82

You are tossing a fair, 4-sided die twice, whose sides are labeled 1, 2, 3, and 4. What is the expected value of the sum of the two throws?

We start by taking the PMF of all of the possible outcomes of the sum of two throws. pX(2) = P(roll 1,1) = 1/16; pX(3) = P(roll 1,2) + P(roll 2,1) = 2/16; pX(4) = P(roll 1,3) + P(roll 2,2) + P(roll 3,1) = 3/16; pX(5) = P(roll 1,4) + P(roll 2,3) + P(roll 3,2) + P(roll 4,1) = 4/16; pX(6) = P(roll 2,4) + P(roll 3,3) + P(roll 4,2) = 3/16; pX(7) = P(roll 3,4) + P(roll 4,3) = 2/16; pX(8) = P(roll 4,4) = 1/16 Then we can take the expected value by multiplying these PMFs by the xi value that they correspond to. Thus, E(X) = (2)(1/16)+(3)(2/16)+(4)(3/16)+(5)(4/16)+(6)(3/16)+(7)(2/16)+(8)(1/16) = 80/16 = 5

A six-sided die is rolled twice. Let X be the sum of the two rolls. What are the possible values that X can take?

X={2,3,4,5,6,7,8,9,10,11,12}