University:
University of Michigan
Course:
MATH 116 | Calculus II
Academic year:

2024
Views:

409

Pages:

2
Author:

banenetta0l

Question 1 Two fair dice are rolled. Let 𝑋 X equal the product of the 2 dice. Compute 𝑃 ( 𝑋 = 𝑖 ) P(X=i) for 𝑖 = 1 , 2 , … i=1,2,…. Answer 𝑋 = 1 ( 1 , 1 ) 𝑝 = 1 36 X=1(1,1)p= 36 1 𝑋 = 2 ( 1 , 2 ) , ( 2 , 1 ) 𝑝 = 2 36 X=2(1,2),(2,1)p= 36 2 𝑋 = 3 ( 1 , 3 ) , ( 3 , 1 ) 𝑝 = 2 36 X=3(1,3),(3,1)p= 36 2 𝑋 = 4 ( 1 , 4 ) , ( 4 , 1 ) , ( 2 , 2 ) 𝑝 = 3 36 X=4(1,4),(4,1),(2,2)p= 36 3 𝑋 = 5 ( 1 , 5 ) , ( 5 , 1 ) 𝑝 = 2 36 X=5(1,5),(5,1)p= 36 2 𝑋 = 6 ( 1 , 6 ) , ( 6 , 1 ) , ( 2 , 3 ) , ( 3 , 2 ) 𝑝 = 4 36 X=6(1,6),(6,1),(2,3),(3,2)p= 36 4 𝑋 = 8 ( 2 , 4 ) , ( 4 , 2 ) 𝑝 = 2 36 X=8(2,4),(4,2)p= 36 2 𝑋 = 9 ( 3 , 3 ) 𝑝 = 1 36 X=9(3,3)p= 36 1 𝑋 = 10 ( 2 , 5 ) , ( 5 , 2 ) 𝑝 = 2 36 X=10(2,5),(5,2)p= 36 2 𝑋 = 12 ( 2 , 6 ) , ( 6 , 2 ) , ( 3 , 4 ) , ( 4 , 3 ) 𝑝 = 4 36 X=12(2,6),(6,2),(3,4),(4,3)p= 36 4 𝑋 = 15 ( 3 , 5 ) , ( 5 , 3 ) 𝑝 = 2 36 X=15(3,5),(5,3)p= 36 2 𝑋 = 16 ( 4 , 4 ) 𝑝 = 1 36 X=16(4,4)p= 36 1 𝑋 = 18 ( 3 , 6 ) , ( 6 , 3 ) 𝑝 = 2 36 X=18(3,6),(6,3)p= 36 2 𝑋 = 20 ( 4 , 5 ) , ( 5 , 4 ) 𝑝 = 2 36 X=20(4,5),(5,4)p= 36 2 𝑋 = 24 ( 4 , 6 ) , ( 6 , 4 ) 𝑝 = 2 36 X=24(4,6),(6,4)p= 36 2 𝑋 = 25 ( 5 , 5 ) 𝑝 = 1 36 X=25(5,5)p= 36 1 𝑋 = 30 ( 5 , 6 ) , ( 6 , 5 ) 𝑝 = 2 36 X=30(5,6),(6,5)p= 36 2 𝑋 = 36 ( 6 , 6 ) 𝑝 = 1 36 X=36(6,6)p= 36 1 Question 2 An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn.) Answer A wins if the first red ball is drawn on the first, third, fifth, or seventh draw. We’ll count the number of times a red ball appears for the first time. (For example, if a red ball is drawn first, there are 9 𝐶 2 9C2 spots where the other two red balls can be placed. To put it another way, there are 9 𝐶 2 9C2 instances in which A wins on the first draw.) 𝐸 ( 1 ) = 9 𝐶 2 E(1)=9C2 𝐸 ( 3 ) = 7 𝐶 2 E(3)=7C2 𝐸 ( 5 ) = 5 𝐶 2 E(5)=5C2 𝐸 ( 7 ) = 3 𝐶 2 E(7)=3C2 When we sum up the number of favorable events and divide by the number of total events: 𝑆 = 10 𝐶 3 S=10C3 𝐸 ( 𝑋 ) : The number of positive events (position of the first red ball) E(X):The number of positive events (position of the first red ball) S: Total number of occurrences (all possible combinations of the balls) S: Total number of occurrences (all possible combinations of the balls) 𝑃 ( 𝐴 wins ) = ( 9 𝐶 2 ) + ( 7 𝐶 2 ) + ( 5 𝐶 2 ) + ( 3 𝐶 2 ) 10 𝐶 3 P(A wins)= 10C3 (9C2)+(7C2)+(5C2)+(3C2) 𝑃 ( 𝐴 wins ) = 0.05833 P(A wins)=0.05833