Please explain how you solved it. It must be done without using software please,
ID: 3295487 • Letter: P
Question
Please explain how you solved it. It must be done without using software please, thank you.
Consider a fair 6 sided dice where one side is labeled 1, one side is labeled 5 and the others are all labeled 0. The die is rolled twice. Define the events A, B, and C as follows: A= {The first die is a 1}, B = {The first die is a 5}, C = {The second die is a 0} a) What is the sample space for this experiment b) What is the probability associated with each of the sample points c) Find the probability of A, B, and C d) Find P(A Union B) and P(A Intersection B) e) Are the events A and B independent, mutually exclusive, or dependent in some other way? f) Find P(A Union C) and P(A Intersection C) g) Are the events A and C independent, mutually exclusive, or dependent in some other way?Explanation / Answer
a) 00, 01, 10, 11, 05, 50, 55, 15,51
b) P(11)=P(55)=P(15)=P(51) = 1/6 x 1/6 = 1/36
P(00) = 4/6 x 4/6 = 16/36=4/9
P(01)=P(10)=P(50)=P(05) = 4/6 x 1/6 = 4/36= 1/9
c) You should be able to get these three without doing all this work (you are looking at one die at a time)
P(A) = P(10)+P(11)+P(15) = 4/36+1/36+1/36 = 6/36 = 1/6
P(B) = P(50)+P(51)+P(55) = 4/36+1/36+1/36 = 6/36 = 1/6
P(C) = P(10)+P(50)+P(00) = 4/36 + 4/36 + 16/36 = 24/36 = 4/6 = 2/3
d) P(AB) = 0 because these two events have no sample points in common
and so P(AUB)=P(A)+P(B)
=1/6+1/6
=2/6
=1/3
e) They are mutually exclusive
f) P(AC) = P(1st die is a one AND 2nd die is a zero) = P(10)=4/36
= 1/9
P(AUC) = P(A) + P(C) - P(AC) = 6/36 + 24/36 - 4/36
= 26/36
= 13/18
g) Independent because P(A|C) = P(AC)/P(C)
= (1/9) / (2/3)
= 3/18
= 1/6 = P(A)
and P(C|A) = P(AC)/P(A)
= (1/9) / (1/6)
= 6/9
=2/3 = P(C)