# Math Help - Conditional probability

1. ## Conditional probability

Imagine you have a wall with a hole.Behind the wall there are 10 targets.On the other side, you have a box with 52 darts.Only one of the darts has the word throw on it.Each target will be showed in the hole in sequence.On every target showed you will pick a dart.If the dart has the word throw, you´ll throw the dart and another dart with the word throw will be put in the box, if not you´ll put it back in the box and you´ll repeat the operation.

if a dart is thrown what is the probability of the target #5 be in the whole ?

2. A = #5 target showing up in the hole
B = Dart labelled 'throw' is picked

By definition:
$P(A \big| B) = \frac{P(A \cap B)}{P(B)}$

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

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.

3. Originally Posted by o_O
A = #5 target showing up in the hole
B = Dart labelled 'throw' is picked

By definition:
$P(A \big| B) = \frac{P(A \cap B)}{P(B)}$

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

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.
If instead of I say that only one dart is thrown , I say that at least one dart is thrown in the 10 attempts , does the calculation change ?