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The question is originally in swedish.

The question translated from swedish to english: A machine gets $exactly$ A, B or C errors with respective probabilities 0.10, 0.07, 0.05. Determine the conditional probability for exactly $A$ errors given that the machine already has exactly one error. The errors are happening independent from each other meaning that machine error 2 doesn't depend on machine error 1 and etc.

Google translate to English:

"A machine gets A, B or C errors with respective probabilities 0.10.0.07 and 0.05. Determine the contingent probability of A error given that the machine has exactly an error. The errors are assumed to occur independently of each other."

My solution:

Let $x_1$ = "The machine gets exactly A errors"

Let $x_2$ = "The machine gets exactly B errors"

Let $x_3$ = "The machine gets exactly C errors"

Let $y_1$ = "The machine has one error"

Let $z$ = "The machine gets exactly A errors given it already has one error"

We use the conditional probability formula: P(z) = P($x_1$ | $y_1$) = $\frac{P(x_1 \cap y_1)}{P(y_1)}$

but we do not know the probability for exactly one error, so how do we continue?

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  • $\begingroup$ "The machine gets $A$ errors" means exactly $A$ errors or at least $A$ errors? $\endgroup$ – BallBoy May 4 '18 at 14:23
  • $\begingroup$ @Y.Forman I think it means exactly $A$ errors. $\endgroup$ – SwedeGustaf May 4 '18 at 14:25
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    $\begingroup$ Can you expound a bit on "The errors are happening independent from each other"? $\endgroup$ – BallBoy May 4 '18 at 14:34
  • $\begingroup$ Can any of $A$, $B$, $C$ be $0$? $\endgroup$ – paw88789 May 4 '18 at 14:36
  • $\begingroup$ @Y.Forman I think they mean that error 2 doesn't depend on error 1, error 3 doesn't depend on error 2 and etc. Just like a result from a dice throw doesn't depend on earlier dice throws. $\endgroup$ – SwedeGustaf May 4 '18 at 14:46
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I don't think the problem is clearly stated. But my best guess is that $A$, $B$, $C$ represent distinct positive integers. And that these are the only possible (nonzero) number of errors. I.e., the number of errors is $A$, $B$, $C$, or $0$.

With this interpretation, then the probability of at least one error is $.10+.07+.05=.22$.

Then the probability of $A$ errors given at least one error is $\frac{.10}{.22}=\frac{5}{11}$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Jyrki Lahtonen May 5 '18 at 18:58

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