This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH242/STAT242/STAT542 Theory of Statistics Solution to Problem Set 1 prepared by Ankur Sharma 1 (Chapter 1 Problem 60) Let E i be the event that an item is produced in the i th shift. Let D be the event that the item is defective. We are given P ( E 1 ) = P ( E 2 ) = P ( E 3 ) = 1 / 3 We are also given P ( D  E 1 ) = 0 . 01, P ( D  E 2 ) = 0 . 02, P ( D  E 3 ) = 0 . 05 So, P ( D ) = ∑ P ( D  E i ) P ( E i ) = (1 / 3) * (0 . 01 + 0 . 02 + 0 . 05) = 0 . 08 / 3 i.e., 8 3 % of items are defective. Also, probability that an item was produced in the 3rd shift given that it was defective is P ( E 3  D ) = P ( E 3 ∩ D ) P ( D ) = P ( D  E 3 ) * P ( E 3 ) P ( D ) = . 05 * (1 / 3) (0 . 08 / 3) = 5 / 8 2 (Chapter 1 Problem 77) Probability of hitting the bull’s eye for the player in any trial = p = 0 . 05 So probability of not hitting in any trial =1 p = 0 . 95 Thus, probability of never hitting in n trials = (1 p ) n = 0 . 95 n and the probability of hitting at least once in n trials = 1 probability of never hitting = 1 . 95 n If this is 0 . 5 then, 1 . 95 n = 0 . 5 or n = log 0 . 5 log . 95 = 13 . 51 n has to be an integer, so for n = 14, the probability of hitting atleast once crosses 0 . 5 3 (Chapter 2 Problem 25) Probability of being dealt royal straight flush = p = 1 . 3 * 10 8 total number of hands seen = n = 100 * 52 * 20 = 104000 So N , the number of royal stright flushes dealt, follows a Binomial(n,p) distribution...
View
Full
Document
This note was uploaded on 02/02/2011 for the course STAT 542 taught by Professor Lisazhang during the Spring '11 term at Columbia.
 Spring '11
 LisaZhang
 Statistics

Click to edit the document details