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I have a hard time understanding this probability example. Suppose behind a door, there is either X, Y or Z.

Your told: The probability of X being behind the door is 50%. The probability of Y being behind the door is 25%. The probability of Z being behind the door is 25%.

Someone goes and opens the door and tells you that Z is not behind the door. What is the probability that Y is behind the door with this new information?

How would one answer this question? And how would you even begin to tackle it?

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  • $\begingroup$ @MohammadZuhairKhan that comes out to be $33.3\%$ $\endgroup$
    – Phil H
    Commented May 12, 2019 at 18:43
  • $\begingroup$ @PhilH absolutely! I am so sorry, I was thinking one thing and typing the other. My bad! $\endgroup$
    – For the love of maths
    Commented May 12, 2019 at 22:01

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After it is revealed that $Z$ is not behind the door, the ratio of probability of $X$ being behind the door to the probability of $Y$ being behind the door does not change. Originally we have $\frac{P(X)}{P(Y)} = 2$, so now we have $P(X) = \frac{2}{3}$ and $P(Y) = \frac{1}{3}$, since the probabilities must sum to $1$.

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