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