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A coin factory produces coins with a probability of getting “tails” X that has a continuous distribution with expectation $0.5$ and standard deviation $0.01$. A person chooses a random coin that is produced in the factory. Suppose that event A is dependent on the value of continuous random variable X . In this case the total probability formula to calculate the probability of event A is $ \int_{-∞}^{∞} P(A|X=x) f_x(x) \,dx $

We define the event: A - “tails” was obtained in a single toss of the coin,

Why $P(A|X=x) = x$ ? It does not make sense because X represents the chance to get "tails" in each coin, so $P(A|X=x)$ represents the chance to get tails in a single toss we have given that $X=x$ , but $X=x$ should represent a value of distribution function (which is unknown because we cannot identify how X distribute), isn't ?

DanielMa
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    Not following. Saying that $X=x$ means that the probability of getting tails is $x$, right? So your question asks for the probability of getting tails, conditioned on the fact that the probability of getting tails is $x$. It doesn't matter how $X$ was chosen, if we know it is equal to $x$. – lulu Mar 23 '24 at 15:05
  • I don't know, it was a question in my exam and did not understand the wording at all. As you say ,"X=x means that the probability of getting tails is x , right?" it should not be $P(X=x)$ ? – DanielMa Mar 23 '24 at 15:12
  • Yes, the probability of getting that value for $X$ is irrelevant, we're told that this is the value. That's all that matters. – lulu Mar 23 '24 at 15:20
  • Yes, but also we know that for every X with continuous distribution $P(X=x) = 0$ , so I completely don't understand what they want in that question – DanielMa Mar 23 '24 at 15:42
  • You should ask your instructor how they expect you to handle conditioning on an event with probability $0$. In this case, I would say the answer was $x$, for the reasons I gave. There is also a case to be made for saying it is ill-defined, since, if you write out the definition of conditional probability formally, you get $\frac 00$. – lulu Mar 23 '24 at 15:46
  • Just to illustrate, suppose $X$ is uniform on $[0,1]$. Then, of course, $P(X=.5)=0$. But I would still argue that $P(X=.5,|, X=.5)=1$ despite the apparent division by $0$. – lulu Mar 23 '24 at 15:47
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    here is a discussion on various ways to broaden the notion of conditional probability to (sometimes) permit conditioning on events of probability $0$. Again, you should ask your instructor how they meant for you to handle it. – lulu Mar 23 '24 at 15:52

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