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: UNIVERSITY OF VICTORIA MATHEMATICS 151, SECTION A02, FALL 2008 SITDOWN PRACTICE TEST 2, NOVEMBER 19 Question 1. You roll a fair 4 sided die twice and multiply the numbers showing on the top face of each roll. What is the expected value of this experiment? (The numbers on the faces of a 4 sided die are 1 , 2 , 3, and 4). (A) (B) 1 (C) 2 (D) 3 (E) 4 (F) 5 (G) 6 (H) 7 (I) 8 (J) 9 Solution: The possible results of two rolls can be summarized by the following table: 1 2 3 4 1 1 2 3 4 2 2 4 6 8 3 3 6 9 12 4 4 8 12 16 Each of these occurs with probability 1 16 , and so summing up we find that the expected value is 100 16 = 6 . 25. So the answer is (G). Question 2. There is a game where the player flips a fair coin 6 times. If they get exactly 2 heads they must pay $4. If they get exactly 3 heads, they receive $4, and if they get exactly 4 heads they receive $12. Otherwise, they get and pay nothing. How much should they expect to win on average by playing this game repeatedly? (A) $0 . 50 (B) $0 . 00 (C) $0 . 50 (D) $1 . 00 (E) $1 . 50 (F) $2 . 00 (G) $2 . 50 (H) $3 . 00 (I) $4 . 00 (J) $6 . 00 Solution: Let X be the random variable that counts heads. This is a binomial random variable with n = 6 trials, and the probability of a heads is p = 1 2 . Thus, the expected winnings for the player in this experiment is $4( P ( X = 2)) + $4( P ( X = 3)) + $12( P ( X = 4)) = $4 6 2 (0 . 5) 2 (0 . 5) 4 + $4 6 3 (0 . 5) 3 (0 . 5) 3 + $12 6 4 (0 . 5) 4 (0 . 5) 2 = $4(15) + $4(20) + $12(15) 2 6 = $200 64 $3 . 125 ....
View
Full
Document
This note was uploaded on 02/05/2011 for the course MATH 377 taught by Professor Stephenlang during the Spring '11 term at University of Victoria.
 Spring '11
 stephenlang
 Math, Calculus

Click to edit the document details