If the probabilities that three children X, Y ,Z will get a ticket for a football game are 0.4, 0.3, 0.2 respectively, calculate the probability that, (Assume that the events of X,Y,Z are independent)
1) None will get the ticket
2) Only one will get the ticket
3) At least one will get the ticket
4) All will get the ticket
Are my answers correct
1) (3 - P(A,B,C))/3 = (3 - (0.4+0.3+0.2))/3 = (3-0.9)/3 = 2.1/3 = 0.7
2) Least probability if of Z (0.2/3 = 0.067)
Max Probability is of X (0.4/3 = 0.13)
So the probability that ONLY one will get ticket will be in the range of 0.67 to 0.13.
3) ???
4) P(A,B,C) = P(A) + P(B) + P(C) = 0.4+0.3+0.2 = 0.9/3 = 0.3