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So, I understand what they are but not perfectly and I'd like to.

Take this example. We flip a fair coin. If it shows Heads, then we flip a fair 6-sided die, else, flip a 3-sided die.

Let A be the event an even number is rolled. Find $P(A|H)$

When I approach these questions, I never know whether to convert it into $P(A ∩ H)/P(H)$ or just answer it using the conditional. I'm not even sure you can without converting it but this is precisely why I'm confused. How do you approach conditionals? Even more complex ones

Edit: what's with the down votes? Did I say something wrong

Tingo Hugo
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  • As long as we stay away from zero-probability events, it’s hard to conceive when $P(A \cap H)/P(H)$ would be a wrong approach. What do you mean by “just answer it using the conditional”? Can you edit the question to give an example of how you would do that? – David K Dec 28 '22 at 22:59
  • @DavidK So for this example, when I first approached it I thought of P(A|H) just as the probability of seeing no odds when we flip the 6-sided die. I assumed that flipping the 6 sided die already took into calculation that I flipped a head and that's where I'm confused. To answer P(A|H), can I just work out the probability of seeing no heads when I'm flipping the 6 sided die? I dont know if that makes sense – Tingo Hugo Dec 28 '22 at 23:09
  • When the events are independent, P(A|H)=P(A). – herb steinberg Dec 28 '22 at 23:17
  • @herbsteinberg right but A and H aren't independent. – Tingo Hugo Dec 28 '22 at 23:21
  • Right, in this case you can take a bit of a shortcut because if you know the coin was heads, it’s obvious what the probability of getting an even number is, because you know the die must be the six-sided one. So yes, I suppose that is how we usually handle a case like that. A problem like finding $P(H\mid A)$ is more complicated, but of course that’s a different problem. – David K Dec 28 '22 at 23:49
  • These things get very unintuitive when $P(H)=0$, but if you exclude that case then intuition ought to hold up. You can always default to the definition, as you mentioned, to confirm. What do you imagine the answer is in this case? Should say that I am not sure what your event $A$ actually is here. I am interpreting it as "the die comes up with an even number" but I am not sure I am right. – lulu Dec 29 '22 at 01:41
  • @DavidK how could you even work out P(H∣A)? Does {A,A'} partition the sample space for H? – Tingo Hugo Dec 29 '22 at 17:26
  • @lulu Yes I just phrased it very poorly. Well, at first glance I think it's 3/6 but I don't understand how you would work out the answer using P(A∩H)/P(H). Isn't P(A∩H) just rolling the 6-sided die and seeing an even number? But then (1/2)/(1/2) = 1 so I don't get it – Tingo Hugo Dec 29 '22 at 17:30
  • Again, I am not sure what your event $A$ is. As I mentioned, I am guessing that you mean "an even number comes up" but I am not sure I am right. Perhaps you could edit your post to clarify. – lulu Dec 29 '22 at 17:45
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    Anyway, if my guess is correct, then $A\cap H$ is the event "you get Heads AND the $6$ sided die comes up even" and that is a probability $\frac 14$ event. – lulu Dec 29 '22 at 17:46
  • But how do you work that out? Do you know A and H are independent? If so, P(A) * P(H) I understand – Tingo Hugo Dec 29 '22 at 18:11
  • No, they are not independent. The probability of getting an even roll increases if you get $H$. Given you got $H$, there are $6$ equally probable things that could happen next, half of which feature an even roll. – lulu Dec 29 '22 at 18:32
  • @lulu Wait so how is it 1/4 I'm so confused EDIT: oh P(A∩H) = 1/2 so P(A∩H)/P(H) = 1/4 – Tingo Hugo Dec 29 '22 at 18:41
  • Let's just list the events. There are $9$ possible events but they are not equally probable. The $6$ events $(H,1), \cdots, (H,6)$ collectively have probability $\frac 12$. As these are equally probable, each has probability $\frac 1{12}$. There are three events contributing to $A\cap H$, namely, $(H,2), (H,4), H(6)$. These collectively have probability $3\times \frac 1{12}=\frac 14$. – lulu Dec 29 '22 at 18:48
  • Of course the three events $(T,1), (T,2), (T,3)$ also have, collective, probability $\frac 12$ so each of those has probability $\frac 16$, but these are not relevant to the computation of $P(A\cap H)$. – lulu Dec 29 '22 at 18:49
  • That gave me a new approach to these questions, I never thought of writing out the possible outcomes as (H,1) thank so much – Tingo Hugo Dec 29 '22 at 18:56
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    I find that technique to be frequently useful, especially when there is confusion. Break the thing down to its component outcomes. Generally (though not always) it's relatively easy to sort out the component outcomes. – lulu Dec 29 '22 at 19:05
  • Of course ${A,A^C}$ is a partition of the sample space — any event and its complement will do that — but it’s not a particularly useful partition. To compute $P(H\mid A)$ I would usually apply $P(H\cap A)/P(A).$ The calculation of $P(A)$ itself takes multiple steps. – David K Dec 29 '22 at 20:02

1 Answers1

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To summarize the discussion in the comments:

There are $9$ possible outcomes to this game, though they are not equally probable.

They split into two groups, according to the coin toss. Within each group, the outcomes are equally probable.

The first group consists of six outcomes, namely: $$(H,1),\,(H,2),\,(H,3),\,(H,4), \,(H,5), \,(H,6)$$

using what I hope is obvious notation.

Now, this group collectively has probability $\frac 12$ since these are all the outcomes attached to tossing $H$ and that is a probability $\frac 12$ event. And these outcomes must be equally probable as the die is assumed to be fair. Thus each of these six outcomes has probability $\frac 1{12}$.

Of course, the other group, $(T,1),\,(T,2),\,(T,3)$ also has, collectively, probability $\frac 12$ so each of those three outcomes must have probability $\frac 16$, but we do not need this to address your question.

There are three outcomes that comprise the event $A\cap H$, namely: $$(H,2),\,(H,4),\,(H,6)$$

As each of these outcomes has probability $\frac 1{12}$, we see that $$P(A\cap H)=3\times \frac 1{12}=\frac 14$$

It follows that $$P(A\,|\,H)=\frac {P(A\cap H)}{P(H)}=\frac {1/4}{1/2}=\frac 12$$ and we are done.

lulu
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  • After learning from this answer, I managed to work out P(H|A) using the similar steps and also implementing the law of total probability. – Tingo Hugo Dec 30 '22 at 18:18