Question

Problem 6 Probability + Counting (3x 3 x 2 18 pts) An urn A coutains tem labeled balls whie each label containsa tumber, rangleg from 2.to 10. An urn B contains five labeled balls while the 1.2,to 5 (a) Two balls are drawn, one froan A and one from B. What is the sample space? What is the probability thst the sum of the labels on the balls is odd? What is the probahility that the sam of the labels on the balls is 9 number is ranging from (b) Two balls are drawn one after the otber without replacetnent and the order matters from urn A. What is the sample space? What is the probability that the sum of the labels on the balls is odd and the first ball number mast be greater than the 2nd ball number)? What is the probability that the sum of the labels on the balls is 9? MAT 115 SPRING 2019 ) Two balls are drawn from urn B one after the other with replacement and the order matters. What is the sample space? What is the probability that the sum of the labels on the balls is even? What is the probabálity that the sum of the labels on the balls is 10? MAT 115 SPRIYC 2n

Answer 1

Problem 9

An urn A contain 10 numbered balls and urn B contain 5 numbered balls.

(a) Here possible value of A is {1,2,3,4,5,6,7,8,9,10}

possible value of B is {1,2,3,4,5}

sample space = {(1,1),(1,2),(1,3),(1,4),(1,5)...................(10,1), (10,2),(10,3),(10,4),(10,5)}

Total sample space points = 10 * 5 = 50

Here first we have to find sum of the labels on the balls is odd. We will find the total sample space points for which sum is odd.

sum is 3 = (1,2), (2,1)

sum is 5 = (1,4), (2,3), (3,2), (4,1)

sum is 7 = (2,5), (3,4), (4,3), (5,2),,(6,1)

sum is 9 = (3,6), (4,5), (5,4),(6,3), (7,2), (8,1) = 6

sum is 11 = (6,5), (7,4), (8,3), (9,2), (10,1)

sum is 13 = (8,5), (9,4), (10,3)

sum is 15 = (10,5)

Total such values = 26

Pr(sum is odd) = 26/50 = 0.52

Pr(Sum on the balls is 9) = 6/50 = 0.12

(b) Two balls are drawn from urn A one after another. Here order matters.

so there are 10 possibliites for first pick and 9 possiblities for second pick so total possibliites = 10 * 9 = 90

Sample space = {(1,2), (2,1) , (1,3), (3,1)...............}

Here condition is sum of the balls is odd and first number is greater than the second number.

So here possible combinations are given below

{(2,1), (3,2), (4,3),(4,1), (5,4),(5,2), (6,5), (6,3), (6,1), (7,6),(7,4), (7,2), (8,7), (8,5), (8,3), (8,1), (9,8), (9,6), (9,4), (9,2), (10,9), (10,7), (10,5), (10,3), (10,1)}

Total = 25

So, Probability = 25/90 = 5/18

Total combination when sum of the labels are 9 = {(1,8), (8,1), (2,7), (7,2), (3,6), (6,3), (4,5), (5,4)}

Probability that sum of the labels are 9 = 8/90 = 4/45

(c)

Two balls are drawn from urn B one after another. with replacement. Here order matters.

so there are 5 possibliites for first pick and 5 possiblities for second pick so total possibliites = 5 * 5 = 25

Sample space = {(1,1),(1,2), (1,3), (1,4), (1,5), (2,1), (2,2),(2,3), (2,4), (2,5), (3,1), (3,2), (3,3),(3,4), (3,5), (4,1), (4,2), (4,3),(4,4), (4,5), (5,1), (5,2), (5,3), (5,4),(5,5)}

Here we have to find the probability that sum of the labels on the balls are even

So here possible combinations are given below

{(1,1), (1,3), (1,5), (2,2), (2,4), (3,1),(3,3), (3,5), (4,2),(4,4), (5,1), (5,3), (5,5)}

Total = 13

So, Probability = 13/25

Total combination when sum of the labels are 10 are = {(5,5)}

probability that sum of the labels are 10 = 1/25