ECON 1203 Tutorial Sample SolutionsSemester 1 2016 Weeks 3 and 4 1. (a) Explain what it means to say that two probabilistic events in a sample space are mutually exclusive of one another. If two events – let’s call them A and B – are mutually exclusive, then it means that they do not have any simple events in common: i.e., that the simple events that combine to make up A have no elements in common with those that make up B. (b) Explain what it means to say that two probabilistic events in a sample space are independent of one another. When two events are independent of one another, it means that the effect of conditioning on the occurrence of one of them has no effect of the marginal probability of the other: i.e., Pr(A/B) = Pr(A). (c) Why can two events not at the same time be both mutually exclusive and independent of one another? Because if A and B are mutually exclusive, then Pr(A and B) = 0, whereas if they are independent, Pr(A and B) = Pr(A)*Pr(B) ≠ 0. 1 212 0.024 0.208 0.072+0. 3 defective ones.192 0. Price category Payment Cash Credit card Debit card Under $20 15 9 18 $20-$100 11 53 52 Over $100 6 38 48 Convert the table to a joint distribution.472 1 (a) What is the probability that an item is under $20? P(Under $20) = 0. cash): P($20-$100|cash) = 0. in fact.344 This implies dependence. In a small batch of 20 manufactured widgets.168 $20-$100 0. selected without replacement.152 0. decide to examine a sample of 3 widgets. 2 .044 (c) What is the probability that people pay for an item that is at least $20 by credit? P(≥$20 and credit) = 0. Express each of the following questions in terms of probability statements. as quality control officer for the company making the widgets.168 (b) What is the probability that an item with a price tag of $43 is paid for in cash? P($20-$100 and cash) = 0.212 + 0.464 3.128 0.593 (e) Are price and means of payment independent? One way to check is to compare the marginal distribution of price with the conditional distribution of price given a particular payment type (say.2. A department store wants to study the relationship between the way customers pay for an item and the price of the item.036 0. You.464 Over $100 0.152 = 0. there are.364 (d) If somebody used a debit card to pay for an item. what is the probability that the item was less than $100? P(<$100|debit) = (0.072 0. 250 transactions are recorded and the following table is formed. ≠ P(($20-$100) = 0. (a) Use a probability tree to evaluate the probability distribution of the number of defectives sampled.044 0. to see how many defective ones are selected.472 = 0.4 0.208)/0. and then solve: Joint distribution: Price category Means of Payment Cash Credit card Debit card Marginal Under $20 0.06 0.368 Marginal 0. (c) This is not an example of the Law of Large Numbers. the 8 probabilities of defective and non-defective are. such as El Nino/La Nina. At face value. Work through problem 44 on page 203 of Sharpe (Chapter 5). Since draws are made independently each time. the number of defectives drawn in a sample of 3 without replacement. The manager of a factory has determined from past experience that X. 2/18. From the lower first branch. 5.10 0.25 0. is x 0 1 2 3 P(X = x) 680/1140 408/1140 51/1140 1/1140 Σ P(X=x) = 1 (b) How would your answer change if the sampling were done with replacement? The resultant probability distribution is now x 0 1 2 3 P(X = x) 4913/8000 2601/8000 459/8000 27/8000 Σ P(X=x) = 1 4. (b) The radio announced is referring to the so-called “law of averages”. From the nodes at the end of the 4 second branches.41 0. There are 14 boxes left. 1/18. the weather is not more likely to be bad in the winter because of a few sunny days in autumn. Work through problem 16 on page 200 of Sharpe (Chapter 5). the player’s frustration grows when his performance starts to “streak” poorly.15 and not defective is 0. 17/18. causing him to work harder to get a hit the next time). (c) Standard statistics says that there is no such thing as being “due for a hit” (the statement being based on the so-called “law of averages”): the batter’s chance for a hit should not change based on recent successes or failures. the relevant probability distribution of X. of which 10 are men’s bikes and only 4 are women’s bikes.06 Calculate the following: 3 .g. The only way that such a statement could be justified would be through some sort of psychological or physiological process that causes repeated performances to be correlated in some way (e. 16/18. The only way that such a statement could be justified would be with reference to some type of global weather event that affects weather patterns in both winter and summer. 6. (b) Her thinking is correct. has the following probability distribution: x P(X = x) 0 1 2 3 4 0. the probability of a defective is 3/19 and of a non-defective is 16/19. 2/18. Work through problem 18 on page 200 of Sharpe (Chapter 5).18 0. From the upper of these branches at the next node the probability of defective being selected is 2/19 and non-defective is 17/19. the number of repairs required to machines in her factory on any one day. 3/18 and 15/18.The tree is of the obvious kind with the first branch from a branch where the probability of defective is 0.85. respectively. which is a mistaken belief that probability will compensate in the short term for odd occurrences in the past. The box selections are not independent of each other: the boxes are not put back into the choice set once they are opened.. 16/18. 7. 10 = 0.06 = 1.15)2 × 0.25 + 2 × 0.15)2 × 0.2 4 . Decisions to do with how many spare parts to have in stock. A human resource manager records the daily numbers of errors of two randomly selected tellers. (g) Describe at least one business decision the manage might face that would be impacted by the information in the table of conditional probabilities.15)2 × 0.31 0.18 + 3 × 0. with the key aspect that the decisions affected must have to do with a response to information about a repair call being observed (since these are conditional probabilities): for example.25 + (2 − 1.5075 (e) What is the conditional probability distribution of X.2 0.17 0.15 Var(X) ???(?) = ?(? − ?)2 = ? 2 = � (? − ?)2 ?(? = ?) ??? ? = (0 − 1.42 0.06 = 1. Suppose that the daily number of errors a randomly-selected bank teller makes is denoted by X and follows the distribution given in the table below.(a) P(1 < X < 4) P(1 < X < 4) = P(X=2)+P(X=3) = 0. are obvious ones.1 + 4 × 0. As the selection is random.18+0. and how many repair technicians to have on call at any one time.41 + 1 × 0.15)2 × 0. many different ideas are plausible. Again. whether additional spare parts or additional technicians must be sourced immediately once it becomes known that a repair call has been made. X1 and X2 are independent and follow the same distribution as X. Denote the associated random variables by X1 and X2.94 (c) E(X) ?(?) = ? = � ??(? = ?) ??? ? (d) = 0 × 0. ? +? The manager then computes the sample mean ?� = 1 2 2 where the sample size is n = 2.28 (b) P(0 ≤ X ≤ 3) P(0 ≤ X ≤ 3) = P(X=0)+P(X=1)+P(X=2)+P(X=3) = 1-P(X=4) = 0.18 + (3 − 1. X 0 1 2 P(X = x) 0.10 + (4 − 1.10 Describe at least one business decision the manager might face that would be impacted by the information in the original table of unconditional probabilities. 8. conditional on some positive number of repairs taking place? x P(X = x|x>0) (f) 1 2 3 4 0.6 0. Many different ideas are plausible here.15)2 × 0.41 + (1 − 1. 1 1 0. ?2 ) = 0 because X1 and X2 are independent. We know the possible values for the mean are 0. Compare these with the result from (a) and comment. and the variance of ?� is the variance of X divided by 2 (the sample size).2 2 0. ?2 Probability 0.36 0. 2.04 2.0 ½ 0.6.2 1 0. Hint: you will find it useful to note that ???(?1 . Now we need to assign probabilities to each outcome to produce the probability distribution for the sample mean. ?2 = 0) = 0.12 1. This simplifies the evaluation of the � variance of the random variable ?. (This is known as the sampling distribution of ?�).1 3/2 0. For example. ½. 3/2. ? +? 1 1 ?(?�) = ? � 1 2 2 � = 2 ?(?1 ) + 2 ?(?2 ) = 0. ?(?1 ) = 0. ½.12 1.6 = 0.32 2 4 4 The means of ?� and X are the same. Hence list the probability distribution of ?� for samples of size 2. If n=2 then the possible values for the mean are 0.04 The required probability distribution is therefore: 5 .0 ?� 0 0. so is ?�.12 0. 2.6 ???(?�) = ??? � ?1 + ?2 1 1 � = ???(?1 ) + ???(?2 ) = 0.6 × 0.64 The mean and variance of X2 are the same because they have identical distributions. (d) Find the possible values that ?� may take.(b) Find the mean and variance of X1.04 2. 1.0 1 0.36 The following table lists all possible outcomes and their associated probabilities: ?1 . 3/2. ?(?� = 0) = ?(?1 = 0. (c) Since X1 and X2 are random. 1.1 ½ 0. ???(?1 ) = 0.04 1. Explain why we do not need to find the mean and variance of X2 once we know those of X1.12 2. Find the mean and variance of the random variable ?�.2 3/2 0. which means n multiplied by every positive integer smaller than itself. Passing each course is assumed to be independent of passing other courses.65 × 0.08 0. A=0 (fail A) & A=1 (pass A) B=0 (fail B) & B=1 (pass B) C=0 (fail C) & C=1 (pass C) (c) What is the probability that this student passes exactly two courses? Express this question in terms of probability statements. Her chances of passing each course are 0.36 0. 0.6)2 . Let X be the number of heads in 4 tosses of a fair coin.74 (e) How reasonable is the assumption of independence? Independence is likely to be an unreasonable assumption. 1 If n=3.5 × (1 − 0. Also recall the combinatorial formula for the number of ways of selecting x from n distinct objects (Sharpe page 193): Cxn = ?!/(? − ?)! ?!. 4.28 0.5 = 0.8 × 0.8 × 0. 2/3. 10. and the combinatorial formula is used to account for the fact that the teller who makes 1 error can be the first. ? �?� = 3� = ?13 (0. Answer the following: (b) Define a random variable for each course outcome.8 × 0. respectively.8.2.65 × (1 − 0.8) = 0. 1.465 (d) What is the probability that this student fails at least one course? Express this question in terms of probability statements. note that the mean can only be 1/3 if two tellers make no errors and the remaining one makes 1 error.?̅ ?(?� = ?̅ ) (e) 0 1/2 1 3/2 2 0.04 Examine briefly what would happen if n =3.5 × (1 − 0. P(failing at least one course) = 1 – P(passing all courses) = 1 − 0. the possible values are 0. B and C. the second or the third sampled teller. 5/3.5. and then solve.65 × 0. we get a finer grid of values between the extremes of 0 and 2. being motivated. Let’s call them courses A. (b) What is the probability distribution of X? 6 . 4/3. working hard.e.24 0. In combinatorial form. you will need to use the idea of a factorial of an integer n. being of high academic ability.. So. 0. Results are likely to be dependent (strong positive association) because most of the variability in course outcomes across students is due to idiosyncratic factors about the student him/herself – i.65) + 0. and 0. for example. As n increases. 9.65.5) + 0. and then solve. 2. 1/3. P(passing two courses) = P(pass A & B but fail C)+ P(pass A & C but fail B)+ P(pass C & B but fail A)=0. 3! = 3 × 2 × 1 = 6. To understand this. A student has enrolled in three courses in this semester. labelled ?!. …? For this last sub-question. The importance of these factors means that there is strong within-student correlation of marks in different courses. 0625 The required probability distribution becomes: ? ?(? = ?) 0 1 2 3 4 0.0625 0. 1.25 0. Plugging in.or 4.12 P( Y = y) (e) 0.X can take on values 0.375 +3×0.0625+(-1)2×0.= P(HHHH) = (0.375 12 0. Now we need all possible combinations that will produce each of these outcomes.5)4 = 0.. when X=0. you lose 12.25 + 2×0.0625 -4 0. The general formula for determining Y from X is Y = 5X – 3*(4-X).25 4 0. 2. Hence: y . why not? Directly from the formula given in part (c).25+0+(1)2×0. 3.5 and we’re assuming outcomes are independent P(TTTT) = P(HTTT) = ….) (TTTT) [4C0=1] (HTTT) (THTT) (TTHT) (TTTH) [4C1=4] (HHTT) (HTHT) (HTTH) (THHT) (THTH) (TTHH) [4C2=6] (THHH) (HTHH) (HHTH) (HHHT) [4C3=4] (HHHH) [4C4=1] nCk Value of X 0 1 2 3 4 Each of these combinations are equally likely because on any toss of a fair coin. and so on.g. Formulate the probability distribution of Y based on the probability distribution of X. (This is the notation used in Sharpe. P(H) = P(T) = 0. Equivalent notation that is sometimes used is ??? .0625 = 2 = (-2)2×0. Let the variable Y denote the winnings from this game.25 +(2)2×0. e. on page 221.25 20 0.0625 What is the expected value of Y? Would you like to play this game? If so.25 0. we have: E(Y) = 5E(X) – 3[4-E(X)] = 10 – 12 +6 = 4 7 . possible combinations over n=4 tosses.375 0.25 +4×0.0625 (c) What are the mean and variance of X? E(X) Var(X) = 0 + 1×0. why? If not.0625 = 1 (d) Consider a game where you win $5 for every head but lose $3 for every tail that appears in 4 tosses of a fair coin. of course. This is unlike games in casinos where expected winnings are negative. 11. meaning the game is biased towards the house.375 +12×. this is not a fair game (since in a fair game.0625 = 4. (The two evaluations. expected returns are zero) but it is biased towards the player. give the same value!) If you play the game enough times you would expect to win $4 per game on average.0625 – 4×0.25 + 4×0. Work through problem 41 on page 234 of Sharpe (Chapter 6). where the latter calculation comes directly from the probability distribution of Y given in the table constructed in part (c). Thus. they should win money.025 +20×0. Notice on any one play of the game you still might lose money and hence someone who is extremely risk averse might not want to play the game even though on average.or E(Y) = -12×0. over many plays of the game. 8 .