MATH1715 Tutee Revision Questions

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eing the “odd-man out” situation, all trials independent, and the probability of a “success” being P. Thus Y ∼ geometric(P). You require E[Y ]. For “fun”, you might like to determine E[Y ] as a function of n. Would you play this game if you went out with a football team (n = 11)? 1 Would you play this game if you were with your <strong>MATH1715</strong> tutees (n ≈ 20)? 2 Question: 10/1/2011 I’m sorry, I know I must be bombarding you with emails. I can’t work out whether this question is telling me conditional probabilities or not. Very ambiguous. On a multiple-choice exam with five choices for each question, a student either knows the answer to a question or chooses an answer at random. If the probability that the student knows a correct answer is 2/3, what is the probability that an answer that was marked as correct was not chosen randomly? Thanks for all your help! Response: First write down what you know. P{Knows correct answer} = 2/3 so P{Does not knows correct answer} = 1/3. P{Correct|Knows correct answer} = 1, P{Correct|Does not know correct answer} = 1/5. We are asked to obtain P{Knows correct answer|Correct}. Clearly we need to use Bayes’s theorem. The way the question is put makes it difficult at first glance to determine what is asked. But notice that the phrase “that was marked as correct” is what has happened, it is known or given. What the question is asking is what is the probability the answer “was not chosen randomly”, in other words what is the probability the correct answer was known by the student, given that it was marked as correct. Question: 10/1/2011 An elevator in a building starts with n people and stops at N floors, with n < N. If each passenger is equally likely to get off at any floor, independently of the others, what is the probability that at least two passengers get off at the same floor? Response: You have done this already! Recall Homeworks 1, question 1.6. Remember “a bus starts with 6 people and stops at 10 different stops”? At least two get off at the same floor is the complementary event to no-one gets off at the same floor. Question: 10/1/2011 Suppose that a random variable X has an exponential distribution with parameter λ = 2 and let Y = e −X . What is the set of possible values of the random variable Y , and how do I find the cumulative distribution function FY (y) = P{Y ≤ y}. How would I obtain the probability density function fY (y) and the mean E[Y ]? If Y1 is the first digit in a decimal expansion of Y , what is the probability that Y1 = 6? Response: I strongly doubt any such transformation question will come up! 1 Only if I wanted to waste 15 minutes, assuming each collective toss takes about ten seconds. 2 Only if we could keep awake for three consecutive days! 2
If X ∼ exponential(λ = 2), then recall that P{X ≤ x} = 1 − e −2x so P{X > x} = e −2x . Then FY (y) = P{Y ≤ y} = P � e −X ≤ y � = P{X ≥ − log y} = e 2log y = y 2 . If x values satisfy x > 0, then 0 < e−x < 1 so the range of y is 0 < y < 1. For the probability density function and mean, recall that fY (y) = dFY � (y) , E[Y ] = y fY (y)dy. dy By first digit of Y I assume is meant the first digit after the decimal point. Clearly Y1 = 6 if 0.6 ≤ y < 0.7. Thus P{Y1 = 6} = P{0.6 < Y ≤ 0.7} = FY (0.7) − FY (0.6) 3 . Question: 10/1/2011 Ten people sit down randomly in a row of 10 chairs. What is the probability that Sam will be seated on a row end with Asif beside him? Response: Draw a picture! How many ways can 10 people sit in 10 seats? Here n(S) = 10!, where S denotes the sample space of possible seatings. You now need to determine how many ways can Sam and Asif sit next to each other. Ask Sam to sit down, but not in an end seat. Now have Asif sit next to him. How many ways can Sam and Asif sit next to each other? Now sit the other eight people down. In how many ways can the 10 people sit with Sam and Asif next to each other? Now you need to add on the cases with Sam sat at an end seat. In how many ways can Asif sit next to him? And now have the other eight people sit down. In how many ways can they sit down? (You have already included cases with Asif being in an end seat in the first part above.) Question: 10/1/2011 Five students are randomly picked from a class of 20 students to form a singing group. There are 8 girls and 12 boys in the class. What is the probability that the singing group will contain exactly two girls and three boys? Response: This is a hypergeometric probability. In how many ways can you choose two girls from 8? In how many ways can you choose three boys from 12? Multiplying gives the number of ways n(E) of choosing two girls from 8 and three boys from 12. For n(S), you need to know the number of ways of choosing any five children from 20. Then P{E} = n(E)/n(S). Question: 7/1/2011 We toss a coin repeatedly until we see each side of the coin at least once. What is the mean number of tosses required? 3 Do not worry that I write the cumulative distribution function as right-continuous. Since Y is a continuous random variable P{Y ≤ y} ≡ P{Y < y}. 3 y
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Page 5 and 6: Three runs gave means 2.99884, 3.00
Page 7 and 8: Response: If the machine breaks dow
Page 9 and 10: Let X be a random variable with mea
Page 11 and 12: Hint: Recall kite flying and counti
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