there are 5 cards total, "sample space of 5", there are 3 blue cards, "favorable outcomes of 3".
the chances of getting the first blue will be 3/5.
once the first blue is pulled out, and not put back, the sample spaced changed to 4, and since there are only 2 blue cards, the favorable outcomes changed to only 2.
the chances of getting the 2nd blue card is 2/4 or 1/2.
P( second blue | first blue) = P(first blue) * P(second blue)
[tex]\bf \stackrel{first~blue}{\cfrac{3}{5}}\stackrel{AND}{\cdot} \stackrel{second~blue}{\cfrac{2}{4}}\implies \cfrac{3}{10}[/tex]
