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**o_O** A = #5 target showing up in the hole

B = Dart labelled 'throw' is picked

By definition:

$\displaystyle P(A \big| B) = \frac{P(A \cap B)}{P(B)}$

$\displaystyle P(A \cap B) = {\color{red}\frac{1}{10}} \cdot {\color{blue}\frac{1}{52}} \qquad P(B) = {\color{blue}\frac{1}{52}}$

Only one target out of 10 is labelled #5. Only 1 dart out of 52 is labelled 'thrown'.

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Or if you simply thought about the problem. Since you're given the fact that a dart is thrown, all you have to consider is the probability of hitting the #5 target out of 10 possible targets.