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
Best Answer
No, your answers are wrong.
When events $A$ and $B$ are independent, then $\mathbb{P}(A \cap B) = \mathbb{P}(A) \mathbb{P}(B)$, and $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B) = 1 -(1-\mathbb{P}(A))(1-\mathbb{P}(B)))$.
Let $T_X$ denote the event that child $X$ will get the ticket.
For the first part, none of the kids get the tickets is logically equivalent to (X does not get a ticket) and (Y does not get a ticket) and (Z does not get the ticket). Since these events are independent, $\mathbb{P}( \lnot T_X \land \lnot T_Y \land \lnot T_Z) = \mathbb{P}( \lnot T_X ) \mathbb{P}( \lnot T_Y ) \mathbb{P}( \lnot T_Z) = (1 - \mathbb{P}( T_X))(1 - \mathbb{P}( T_Y))(1 - \mathbb{P}( T_Z))$. So $\mathbb{P}( \lnot T_X \land \lnot T_Y \land \lnot T_Z) = (1-0.4)(1-0.3)(1-0.2) = 0.336$.
For the second question, the logical expression for
Only one will get the ticketis $(T_X \land \lnot T_Y \land \lnot T_Z) \lor (\lnot T_X \land T_Y \land \lnot T_Z) \lor ( \lnot T_X \land \lnot T_Y \land T_Z)$. I will leave to you to work it out, the answer should be $0.452$.For the third question, it is easier to think of the logical opposite, which you have already answer in earlier question.
For the last part, the logical expression is $T_X \land T_Y \land T_Z$, so the probabilities are products $0.4 \times 0.3 \times 0.2 = 0.024$.