Answer:
Step-by-step explanation:
Given that events are as follows:
A: A green ball is drawn from a box with five balls and placed next to the box. B: A red ball is drawn next and placed next to the green one
i.e. drawings of balls are made without replacement
Suppose there are m green balls and n red balls, where m+n =5
Prob of drawing green ball =m/5
After I draw made remaining are n red balls and m-1 green balls
Prob of drawing red ball now =n/4
Prob for drawing one green and one red simultaneously = [tex]\frac{mC1*nC1}{5C2} \\= \frac{mn}{10}[/tex]
Product of the two probabilities = [tex]\frac{m}{5} *\frac{n}{4} \\=\frac{mn}{20}[/tex]
