Answer:
[tex]\frac{9}{55}[/tex]
Step-by-step explanation:
A goblet contains 3 red balls, 2 green balls and 6 blue balls.
Total balls = 3 + 2 + 6 = 11 balls
we choose a ball and then another ball without putting the first one back.
The probability that the first ball will be red = [tex]P_{1}[/tex] = [tex]\frac{3}{11}[/tex]
Second event is done without replacing the first ball so total ball will be 10
The probability that the second ball will be blue = [tex]P_{2}[/tex] = [tex]\frac{6}{10}[/tex] = [tex]\frac{3}{5}[/tex]
P = [tex]\frac{3}{11}[/tex] × [tex]\frac{3}{5}[/tex] = [tex]\frac{9}{55}[/tex]
The probability that the first ball will be red and the second ball will be blue is [tex]\frac{9}{55}[/tex]
