Q1. Describe a sample space representing the results of two coin-tosses. How many outcomes are in your space? Could you also argue for a sample space with a different number of outcomes?
Answer: {HH, HT, TH, TT}, with four outcomes. We could also ignore the order of the results, and say the outcomes are {2 heads, 2 tails, one of each}.
Q2. Specify an algebra π over the sample space you proposed in Q1. Now specify another algebra πβ², with a different cardinality, over that same sample space.
Answer: Using the first sample space, here is one algebra: {β , {HH}, {HT}, {TH, TT} (i.e. "tails-first"), {HH, HT}, {HH, TH, TT} (i.e. "not-HT"), {HT, TH, TT} (i.e. "not-HH"), {HH, HT, TH, TT} (i.e. β€)}. Many other choices are possible, such as {β , β€}.
Q3. What does it mean to say that an algebra "is closed under complement and finite union"?
Answer: π is closed under complement means: βFβπ. (Ξ©-F)βπ, where Ξ©=βπ. π is closed under finite union means: βF,Gβπ. (FβͺG)βπ.
Q4. If a fair coin is tossed three times, how many possible outcomes are there? (Let's distinguish between the order in which we get the different results.) In how many outcomes do we get at least 2 Heads?
Answer: 8 outcomes, with at least 2 Heads in 4 of them: HHT, HTH, THH, HHH.
Q5. For a probability function defined on a finite algebra, is it permitted for the probability of a non-empty event or hypothesis to ever be 0? Why or why not?
Answer: Yes, the axioms say only that H=β entails prob(H)=0, not the converse. Some Bayesians argue that your prior credences for non-empty discrete hypotheses should never be 0, but they don't object to your posterior credences being so.
Q6. If H β G, could it be that prob(H) > prob(G)? Why or why not?
Answer: No, prob(G) = prob(GH) + prob(G β§ not-H). If H β G, then G β© H will be H, so the first operand = prob(H). So prob(G) = prob(H) + prob(G β§ not-H). Since the second operand has to be β₯ 0, prob(G) β₯ prob(H).
Q7. Suppose prob(H) = 1/2, prob(G) = 1/3, prob(H β¨ G) = 2/3. What is prob(HG)?
Answer: prob(H β¨ G) = prob(H) + prob(G) - prob(HG). Plugging in the three values we're given, we solve for prob(HG) = 1/6.
Q8. Suppose H and G are independent, and that prob(H) = 1/2 and prob(G) = 1/3. What is prob(HG)? What is prob(H β¨ G)?
Answer: if H and G are independent, then prob(HG) = prob(H)prob(G) = 1/6. Then prob(H β¨ G) = 2/3, as in previous question.
Q9. If prob(H) and prob(G) are both > 0, and H and G are mutually exclusive, could it be that H and G are probabilistically indepenent? Why or why not?
Answer: No, if they were mutually exclusive, then prob(HG) = 0, so it could not equal prob(H)β’prob(G).
Q10. A six-sided die is biased so that the probability of rolling a 2 is twice that of rolling a 1, that of rolling a 3 is thrice that of rolling a 1, and so on. What is the probability of rolling an odd number?
Answer: prob(1) = x, prob(2) = 2x, etc. prob(odd) = (x+3x+5x)/(x+2x+3x+4x+5x+6x) = 9/21.
Q11. Call a deck of cards "fairly shuffled" when each card then in the deck has an equal chance of being drawn. For a fairly shuffled deck of the standard 52 cards, (a) what is the probability that the first card drawn will be a Heart? (b) What is the probability that it will be a Heart, given that it is Red? (c) What is the probability that it will be Red, given that it is a Heart? (d) What is the probability of Red β Heart, that is, that either the card is a Heart or it is not Red?
Answers: (a) 1/4 for Hearts; (b) 1/2 for Hearts, given Red; (c) 1 for Red, given Hearts; (d) 3/4 for Red β Heart, that is, not-Diamonds.
Q12. Concerning the same card as in the previous question, what is the probability that it is either an Ace or a Heart?
Answer: prob(A but not β₯) = 3/52; prob(A of β₯) = 1/52; prob(β₯ but not A) = 12/52; sum = 16/52. Alternatively, prob(A) = 4/52 + prob(β₯) = 13/52 - prob(A of β₯) = 1/52; sums to the same.
Q13. What is the expected rank of the drawn card? (Let A=1, J=11, Q=12, K=13.)
Answer: (1/13)β’1 + (1/13)β’2 + ... + (1/13)β’13 = 7.
Q14. Suppose that after the first card was drawn, it was not replaced. A second card was drawn from the rest of the (still fairly shuffled) deck. What is the probability of the first card being an Ace, and the second being a Heart?
Answer: (a) prob(first is A of β₯) = 1/52; (b) prob(first is A but not β₯) = 3/52; (c) prob(second is Heart|first is A of β₯) = 12/51; (d) prob(second is Heart|first is A but not β₯) = 13/51. Now, prob(first is A β§ second is β₯) = prob(first is A of β₯ β§ second is another β₯) + prob(first is A but not β₯ β§ second is β₯) = (a)β’(c) + (b)β’(d) = (12 + 3β’13)/(51β’52) = 1/52.
Q15. What is the probability of the second being a Heart, given that the first is an Ace?
Answer: prob(second is β₯|first is A) = prob(first is A β§ second β₯)/prob(first is A). The top value was calculated in previous problem to be 1/52. The bottom value is 4/52. Solution is 1/4.
Q16. What is the probability of neither of those two cards being an Ace?
Answer: (a) prob(first is not A) = 48/52; (b) prob(second is not A|first is not A) = 47/51, since one of the non-A will have been removed; (c) prob(first is not A β§ second is not A) = (a)β’(b) = approximately 85%.
Q17. In point 6 of the notes for this week's readings, we describe a random variable for a scenario involving a coin toss and a spun dial. Assume that the coin and the dial are both fair. What is the expected value of that variable?
Answer: 1/2 prob of value (-1); 1/2 prob of a continuous range of values, all with equal probability, whose average is Ο. So expected value = -(1/2) + (Ο/2).
Q18. Alice has two coins. One is fair, but the other is biased so as to be twice as likely to land Heads as to land Tails. Alice chooses one of her coins (she can tell them apart visually). Bob knows how the two coins work but has no information about which one Alice chose. Before Alice tosses the coin, what should the probability be, according to Bob, that (a) the coin is fair? (b) that the coin will land Heads? (c) What should the probability be, according to Alice, that the coin will land Heads?
Answers: (a) 1/2 is a natural answer; (b) (1/2 fair)(1/2 H given fair) + (1/2 biased)(2/3 H given biased) = 1/4 + 1/3 = 7/12; (c) Either 1/2 or 2/3, depending on which coin she knows herself to have chosen.
Q19. Alice tosses the coin and it lands Heads. What should the probability now be, according to Bob, that the coin is fair?
Answer: prob(fair|H) = prob(Hβ§fair)/prob(H) = prob(H|fair)prob(fair)/prob(H). (This is Bayes' Theorem.) We know that prob(H|fair) = 1/2, and prob(fair) = 1/2, and we calculated the denominator prob(H) as part (b) of the previous question. Solution = 3/7.
Q20. Suppose that prob(H) = 1/2, prob(G|H) = 1, prob(G|not-H) = 1/3. What is prob(H|G)?
Answer: prob(H|G) = prob(Gβ§H)/(prob(Gβ§H)+prob(Gβ§not-H)). prob(Gβ§H) = prob(G|H)prob(H), both of which values we're given, so this is 1/2. prob(Gβ§not-H) = prob(G|not-H)prob(not-H) = prob(G|not-H)(1-prob(H)), and we're given those values, so it's 1/6. Putting it together, we get (1/2)/((1/2)+(1/6)) = 3/4.
Q21. The probability that Clarice will study for the test is 2/3. The probability that she will pass the test given she studies is 1/2. The probability that she will pass given that she doesn't study is 1/6. What is the probability that she will pass the test?
Answer: prob(pass) = prob(pass and study) + prob(pass and not-study) = prob(pass|study)prob(study) + prob(pass|not-study)prob(not-study) = (1/2)(2/3) + (1/6)(1/3) = 7/18.
Q22. Given that Clarice in fact passes the test, what is the probability that she did study?
Answer: prob(study|pass) = prob(pass and study)/prob(pass). The numerator is prob(pass|study)prob(study), which values were supplied, so this is (1/2)(2/3) = 1/3. The denominator we derived in the previous question, that's 7/18. Solution = (1/3)/(7/18) = 6/7.
Q23. Suppose there are three boxes. One has two gold coins, another has two silver coins, and the last has one gold and one silver. You have no information about which box is which. You randomly choose one of the boxes, reach in and pull out a gold coin. What is the probability that the coin remaining in that box is also gold?
Answer: 2/3. See discussion here; or if you want to read more.
Extra credit 24. A fairly shuffled deck has four cards: the Ace and Two of Hearts, and the Ace and Two of Spades. Two cards are dealt to Douglas. There are six possible deals of two cards from a deck of four, all equally likely. In one of them Douglas has both Aces, in five of them he has at least one Ace, in half of them he has the Ace of Spades, and in half he has the Ace of Hearts. Douglas now tells you, "I have at least one Ace." Conditioning on this information (by discarding the outcome where Douglas received no Aces), you now compute the probability that he has both Aces to be 1/5. Next Douglas tells you "I have the Ace of Spades." Conditioning on this new information, you now compute the probability that he has both Aces to be 1/3: if the three deals in which he got the Ace of Spades, he had both aces in one of them. But wait a second. If he told you that he had the Ace of Hearts, you would also end up with the probability that he has both Aces as 1/3. And you already knew he had at least one Ace. It had to be either the Ace of Hearts or the Ace of Spades. Why should finding out its suit make it more likely that he has both Aces? If you were going to raise your probability from 1/5 to 1/3 no matter what suit he revealed, why shouldn't it have been 1/3 all along?
Answer: See discussion here and here.