Answer:
a)P(A1) = P( aaa, abc, acb ) = 1/3
P(A2) = P(aaa, bac, cab ) = 1/3
P(A3) = P(aaa, cba, bca ) = 1/3
b) To verify IF A1, A2 and A3 are independent we can see that
P( A1 and A2 and A3 ) = P( aaa ) = 1/9 is not equal to P(A1)P(A2)P(A3)
Step-by-step explanation:
Based on the given conditions The sample space S can be written as
aaa, bbb, ccc, abc, acb, bac, bca, cab,and cba = 9 most likely outcomes
Defining Ai ;which is the event that ith place in the triple is occupied by a
The probabilities of i=1,2,3 can be computed as
P(A1) = P( aaa, abc, acb ) = 1/3
P(A2) = P(aaa, bac, cab ) = 1/3
P(A3) = P(aaa, cba, bca ) = 1/3
P(A1 and A3) = P(A2 and A3 ) = P ( A1 and A3 ) = 1/9
THIS CAN BE EXPRESSED AS
P(A1)P(A2) = P(A2)P(A3) = P(A1)P(A3)
It can observed that
A1, A2, A3 are pair wise independent of each other
to verify this we can see that
P( A1 and A2 and A3 ) = P( aaa ) = 1/9 is not equal to P(A1)P(A2)P(A3)