MATH 55(31) WD SMITH HOMEWORK #9 DUE TUES 29 APR AT CLASS START --------------------------------------------------------------- ANSWERS PROBLEM #0 (not worth any points :) People seem divided on whether I should grade based on plan A or plan B my tests midterm final homeworks ExtraCred A: 20 30 30 20 + B: 25 25 25 25 + tell me which plan A or B you prefer, or suggest your own weighting scheme if you don't like A or B. PROBLEM #1: Exponential decay of some radioisotope causes it, after 100 years to be 90% gone (only 10% remains). What is the half-life of the isotope? ANS. Thalf = 100 * log(1/2) / log(0.1) = 30.103 years. -30.103/100 SANITY CHECK. 2 = 0.1. PROBLEM #2: If you pick 3 cards from a deck, what is the ODDS that the cards will be (ignoring their suits) Ace, 2, and 3, not necessarily in that order? 64 64 16 ANS. The probability you get A,2,3 is ---- = ----- = ---- = .0028959... 52C3 22100 5525 where 52C3 = 52*51*50/(3*2*1) = 22100 is the number of 3-card subsets of the 52 cards (use C because order does NOT matter), and where the number of ways to have 3 cards in your hand which are A,2,3 (ignoring suits and not counting different orders as different) is 4*4*4 = 64 since the ace could be any of 4 cards (heart, diamond, clubs, spades), the 2 could be any of 4 cards, and the 3 could be any of 4 cards. PROBLEM #3: Suppose your expected remaining life is 60 years, and you are 100% confident you will not live for 130 more years. I claim from those 2 facts alone, you can deduce that the probability you will still be alive in 20 years, is at least 36.36%. Figure out why that is true and explain your reasoning. Then deduce that the probability you will still be alive in 50 years, is at least X -- and please supply the biggest X you can such that you know it is true? ANS. The idea is that, if the probability you will be dead in 20 years were too large, there is no way your life-expectancy could still be as large as 60 years. Let P=Prob(you will still be ALIVE in 20 years). Then even if you then got a free pass to live 130 more years (i.e. 110 more, at that point) and even if, if you are going to be dead in 20 years you knew you'd live 19.999 of them, your life expectancy (of years left starting now) would be AT MOST (1-P)*20 + P*130. I.e. that is an overestimate of your life expectancy. Now since your life expectancy is 60 more years (you've been told) we know (1-P)*20 + P*130 >= 60. That is 20 + 110 P >= 60 110 P >= 40 P >= 40/110 = 4/11 = .3636363636 = 36.4% rounded off. Thus the probability you will be dead in 20 years is at most 63% and you have at least 36% probability of being alive. Similarly (1-X)*50 + X*130 >= 60 so 50+80X >= 60 so 80X >= 10 so X >= 10/80 = 1/8. So there is at least a 1/8 chance you'll still be alive in 50 years. PROBLEM #4: There are 5 urns of candy. They have inside them, respectively, 5%, 17%, 23%, 57%, and 91% blue candies. You reach in blindly and pick one candy from each urn. What is the expected number of blue candies that you will get? ANS. Since expectations add, the expected number of blues is .05 + .17 + .23 + .57 + .91 = 1.93. That is the expected #blues you get in the urn#1 pick, plus the expected #blues you get in the urn#2 pick, +...+ urn#5. PROBLEM #5: You want to have $100,000 in 15 years. So you deposit $X per month into a savings account making 4% APR compounded monthly. What should X be? ANS. USE SAVINGS PLAN FORMULA (version solved for payment) i/n X = PMT = A ----------------- nY (1 + i/n) - 1 We have A=100000, $i = 0.04$, $n = 12$ (since monthly), $Y = 15$. Get $406.35 (rounded off to nearest cent). SANITY CHECK: your total payments are $406.35 * 12 * 15 = 73143.00 which is less than $100000 (due to interest helping you) but not insanely less. PROBLEM #6: Here are some event-pairs. For each pair decide if the events are DISJOINT or OVERLAPPING, and INDEPENDENT or DEPENDENT. ans a) Pick an apple and find it has a worm. It's moldy. DEPENDENT, OVERLAPPING. b) Get killed in a mugging tomorrow. It rains tomorrow. INDEPENDENT, OVERLAPPING. c) Roll a dice and get 6. No, get 5. DEPENDENT, DISJOINT. d) Picking random person, get a female. It rains tomorrow. INDEPENDENT, OVERLAPPING. e) Picking random person, get a female. Get a boy <10 years old. DEPENDENT, DISJOINT. PROBLEM #7. How many orderings of 10 people (half of them male) will have 4 among the first 5 people being male? ANS. 5C4 * 5C1 * 5! = 5 * 5 * 120 = 3000. Because there are 5C4=5 ways to pick the 4 males from the 5, and there are 5C1=5 ways to pick the 1 female from the 5, and then, there are 5!=120 ways to order the resulting 5 people. (This is out of 10! = 3628800 orderings of the 10 people, in all, i.e. a fraction 3000/3628800=5/6048 of all of them. So the chance this will happen is surprisingly small, only about 0.0008.)