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October 1, 2014

October 1, 2014

Posted by **Ana** on Friday, May 11, 2012 at 10:14pm.

- Math -
**MathMate**, Friday, May 11, 2012 at 10:34pmAssuming 6-sided number cubes.

Let the events

A=throwing less than 6

B=one of them is a three

Then P(A∩B)

=|{(1,3),(2,3),(3,1),(3,2)}| / 36

= 1/9

P(B)

=2/6

=1/3

Throwing less than six given one of them is a 3 is therefore the conditional probability of A given B, or

P(A|B)

=P(A∩B)/P(B)

=(1/9) /(1/3)

=1/3

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