Let's say that I have 2 blue balls and 3 green balls. The event is when I draw 2 balls of different colors.
This is generally solved by selecting 1 blue and green ball and dividing it by all the outcomes, i.e: 5c2.
It would look something like this: 2c1*3c1\5c2
But what if I were asked to find the probability where the first ball drawn is blue and the second ball drawn is green? What would I have to do then?
I was thinking that I would have to divide the result I got earlier by 2. Is this correct?
The experiment is that we are drawing two balls without replacement from the bag with $2$ blue balls and $3$ green balls. Let $A_1$ be the event that we draw a blue ball on the first draw and $A_2$ be the event that we draw a green ball on the second draw. Then $$ P(A_1\cap A_2)=P(A_1)P(A_2\mid A_1)=\frac{2}{5}\times \frac{3}{4}. $$