Answer-
The probability that in the box there are 1 red and 8 blue markers is 0.111 or 11.1%
Solution-
In the box all markers are red or blue. There are total of 9 markers.
So the number of possible combination number of red or blue marker is,
[tex]S=[(0,9),(1, 8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1),(9,0)}][/tex]
As at the random drawn, there was a Blue marker, so the condition of (9,0) i.e 9 Red marker and 0 Blue marker is not a case.
So the sample space becomes,
[tex]S=[(0,9),(1, 8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1)}][/tex]
[tex]|S|=9[/tex]
Let us assume that E is the event that in the box there are 1 red and 8 blue markers. So
[tex]E=[(1,8)][/tex]
[tex]|E|=9[/tex]
The probability that in the box there are 1 red and 8 blue markers is,
[tex]P(\text{1 Red and 8 Blue})=\dfrac{|E|}{|S|}=\dfrac{1}{9}=0.111=11.1\%[/tex]
