Joanna ignores her probability professor\'s advice and elects to spend her money
ID: 3206077 • Letter: J
Question
Joanna ignores her probability professor's advice and elects to spend her money playing the lottery. She has $100 available to spend, and she is willing to spend all of it, but decides in advance to buy tickets one by one, stopping as soon as she either has a winning ticket or runs out of cash. Each lottery ticket costs $1 and is known to have one of three outcomes i) no cash value (with probability 99/100), ii) cash value of $10 with probability 1/100 -1/1000, or iii) cash value of $100 with probability 1/1000. Let the RV X denote Joanna's wealth at the end of this process. Please answer the following questions. (a) What is the expected value of Joanna's wealth after this experiment? (b) What is the probability that Joanna loses all her money in this experiment? (c) What is the probability that Joanna loses some money in this experiment? (d) What is the probability that Joanna's wealth stays exactly the same in this experiment? Hints (a) Define 2. The following outcomes w are possible: i) she buys k losing tickets and then a winning $10 ticket for k E 10, ,992, ii) she buys k losing tickets and then a winning $100 ticket for k E 10, ,99) or iii) she buys 100 losing tickets. (b) Assign P(w) or each w E 2; check (with a computer) that Swen P(w) 1. (c) Define the RVX as a function from 2 Z, and find the support Y; you should find I 191 (d) Define E w E 2 X(w) ar or a E X. There are important transitions in the nature of E between 0 and 1, between r 99 and r 100, and between r 109 and 110 (e) Compute p (r) POX POEa P(w) or each a EX (with a computer), and check (with WEEr a computer) that ner p(z) (f) Use the PMF p (p(z), z E X) to answer the questions aboveExplanation / Answer
Ans A) Expected welth after the experiment = (0*0.99+10*(0.01-0.001)+100*0.001)*100 = 100*(0.1-0.01+0.1) = 19
Ans B) Probability of Jenna losing all money = Probability of 100 loosing tickets = 0.99100 (Since individual ticket purchases are independent of each other) = 0.37 (approx)
Ans C) Since she stops as soon as she receives a winning ticket, in this scenareo, she doesn't get a $100 (Otherwise she'll be left with $100 which is either more than or equal to what she spent).
Thus, the probability that she gets less that what she spent = probability that she gets all loosing tickets till 10th try and then getting either a loosing ticket or a $10 ticket thereafter and stopping at a $10 ticket.
= (0.99)10*[(0.009)+(0.99)*(0.009)+(0.99)2*(0.009)+.....(0.99)89*(0.009) ] + (0.99)100 ... This is because as and when there is a $10 ticket with probability 0.009, the game is over. The last term is introduced for loosing all the trials.
= (0.99)10*(0.009)*[1+0.99+0.992.....+0.9989] + (0.99)100
= (0.99)10*(0.009)*(1-0.9990)/(1-0.99) + (0.99)100 = .485+.366 = .851 (Approx, please calculate)
Ans D) This condition is possible if and only if she looses all the first 99 tries and finally purchases a $100 ticket.
The proability = (0.99)99*0.001 = 0.00037
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.