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Problem 1. Professor Lalley has two coins, one fair, the other two-headed. He chooses one of the coins at random and tosses it 3 times

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Problem 1. Professor Lalley has two coins, one fair, the other two-headed. He chooses one of the coins at random and tosses it 3 times in succession. Given that he sees 3 consecutive Heads, what is the conditional probability that he tossed the two-headed coin? Problem 2. There are 100 multiple choice questions in a test. Suppose that independently for each question, James knows the answer with probability 0.75 (in which case, James will always get the correct answer). When James does not know the answer, James guesses randomly out of the 4 candidate choices and thus has probability 0.25 to get the correct answer. Given that James get the correct answers to 85 questions, what is the conditional probability that James knows the answers to exactly 80 questions? Problem 3. Let X and Y be independent random variables with geometric distributions P(X = k) = p(1 -p)*-1 and P(Y = k) =r(1 -r)k-1 for k =1, 2,3, ... Define Z to be the minimum of X and Y. (a) What is the distribution ofZ? (b) What is P(X = k | Z = k)? (c) What is Var(X + Y)? (d) Ifr = p, what is P(X > k + m |, Z =k)? Problem 4. If a fair die is rolledn times in succession, what is the probability that (a) no roll shows the same number as the preceding roll? (b) no roll results in one of the numbers showing on one of the two preceding rolls? Problem 5. Four players sit down to a round of bridge. Each player is dealt 13 of the 52 cards, which we assume were well-shuffled before being dealt. Let be the number of players void. (a) What isEX? (b) What is Var( X) ? Problem 6. Let X and Y be independent random variables withEX = EY = / and VarX = VarY = 02. Write Z = XY . Calculate Var Z in terms off and o. Problem 7. A standard deck consists of 52 cards. Four of the cards are aces. After thorough shuffling, cards are dealt from the deck one by one without replacement. Let 1 be the number of cards dealt till the first ace appears. LetY2 be the total number of cards dealt till the second ace appears. So P(1

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