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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

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Let's look at 1). The solution sketch contains an equation that should not appear anywhere, namely $3-0.9=2.1/3$. And the correct approach is to say that the probability that X fails to get a ticket is $1-0.4$, which is $0.6$. The corresponding failure probabilities for Y and Z are $0.7$ and $0.8$. To find the probability they all fail, by independence we multiply the individual probabilities of failure. The result is $(0.6)(0.7)(0.8)$, which is $0.336$. – André Nicolas Oct 6 '11 at 16:16
up vote 1 down vote accepted

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 ticket is $(T_X \land \lnot T_Y \land \lnot T_Z) \lor (\lnot T_X \land T_Y \land \lnot T_Z) \lor ( \lnot T_X \land \lnot T_Y \land T_Z)$. 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 $0.4 \times 0.3 \times 0.2 = 0.024$.

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$\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B)$ does not hold. Otherwise you could get a probability of 2 by letting A and B both be certain events... – Henning Makholm Oct 6 '11 at 16:18
@HenningMakholm Thank you Henning, I have corrected the embarrassing blunder of mine now. – Sasha Oct 6 '11 at 16:27
so the probability that "Atleast one" will get the ticket is (1 - P(None will get the ticket) = 1-0.336= 0.664 – aadidasu Oct 6 '11 at 17:44

If the probabilities that the kids will get the tickets are $0.4$, $0.3$ and $0.2$ independently, then the probabilities that they won't get the tickets are $0.6$, $0.7$ and $0.8$. The probability that none of them get a ticket is found by multiplying the probabilities that they individually don't get tickets, i.e.

$$P(\textrm{No one gets ticket}) = P(\textrm{X doesn't}) \times P(\textrm{Y doesn't}) \times P(\textrm{Z doesn't}) = 0.6 \times 0.7 \times 0.8 = 0.336$$

Does this help you work out the probabilities for the other questions?

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In that case the probability that all will get the ticket is 0.4*0.3*0.2 = 0.024. But aren't the probabilities of independent events to be added rather than multiplied. – aadidasu Oct 6 '11 at 16:14
@Aa: No, no. Adding probabilities of independent events makes no sense at all. – Henning Makholm Oct 6 '11 at 16:16
On the other hand, we add probabilities associated with mutually exclusive events, which is about as far away from independence as one can be (undergraduates seem to think these are the same for some reason, at least over here.) – guy Oct 6 '11 at 16:19

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