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Toss a coin 3x, find the probability that you get TTT

We have the events:

$A = \{\text{first coin gets tails} \}$

$B = \{\text{second coin gets tails} \}$

$C = \{\text{third coin gets tails} \}$

Clearly the outcome of $A$ affects the total probability.

Or am I getting the definition mixed up?

If the first gets tails, the SECOND/THIRD are unaffected, BUT total probability is affected.

So what is the outcome?

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    $Pr(\{TTT\})=Pr(A\cap B\cap C)=Pr(A)Pr(B)Pr(C)$ since they are independent events. As for why they are independent events, it is usually specified in the problem statement, but for this specific case you can notice that $Pr(A)=\frac{1}{2}=\frac{4}{8}$ since there are four of the eight equally likely outcomes that result in a tail for the first, similarly for $Pr(B)$ and $Pr(A\cap B)$ is $\frac{1}{4}=\frac{2}{8}=\frac{1}{2}\cdot\frac{1}{2}$. Similar calculations for the rest. – 2017-01-23
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    Remember definition of mutual independence: $A,B,C$ are all mutually independent if $Pr(A\cap B\cap C)=Pr(A)Pr(B)Pr(C)$ – 2017-01-23
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    @GerryMyerson, you really spent time noticing that. Wow. – 2017-01-23
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    What do you mean by "total probability"? Probability of ***what***? – 2017-01-23

2 Answers 2

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I think you are mixing things.

3 tosses of a coin are independent. So you can write first all cases. And then see those cases that are favourable.

As 3 rolls of a dice is same as 3 coins are tossed together.

So probability = $\frac18$

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The event {there are three tails} is not independent of event A, but events $B$ and $C$ are. To be explicit, call the event $$D = \{\mbox{there are three tails}\}.$$

Independence of a set of events $E_1,E_2,E_3\ldots$ is defined by when the probability always factors: $$P(E_1,E_2,E_3)=P(E_1)P(E_2)P(E_3)\\P(E_1,E_2) = P(E_1)P(E_2)$$ and so on for any combination of the $E_1,E_2,E_3\ldots$.

Now we can apply that definition. The probability of both $A$ and $B$ is $1/4$ (since that's the probability both the first and the second coin is tails). So we have $$\frac{1}{4} = P(A,B) = P(A)P(B)$$ since $P(A)=P(B) = 1/2.$ This means that $A$ and $B$ are independent. Also note our initial calculation $P(A,B) = 1/4$ is a consequence of our assumption that the coinflips are independent of one another.

However, consider the probability that both $A$ and $D$ occur, i.e. the first coin is tails and all three coins are tails. This just means that all three coins are tails so the probability is $P(A,D) = 1/8.$ However, we have $P(D) = 1/8$ and $P(A)=1/2,$ so $$\frac{1}{8}=P(A,D)\ne P(A)P(D) = \frac{1}{16}.$$ So $A$ and $D$ are not independent