1.)A bag has 7 red marbles, 8 blue marbles and 3 black marbles. What is the probability of picking a blue marble, replacing it, and then picking a red marble?
Answer a. 14/81
b. 7/108
c. 2/27
d. 1/9
2.)A bag has 7 red marbles, 8 blue marbles and 3 black marbles. What is the probability of picking a black marble, replacing it, and then picking a red marble?
Answer a. 2/27
b. 7/108
c. 14/81
d. 1/9

Question

1.)A bag has 7 red marbles, 8 blue marbles and 3 black marbles. What is the probability of picking a blue marble, replacing it, and then picking a red marble? 
Answer   a. 14/81 
  b. 7/108 
  c. 2/27 
  d. 1/9 
2.)A bag has 7 red marbles, 8 blue marbles and 3 black marbles. What is the probability of picking a black marble, replacing it, and then picking a red marble? 
Answer   a. 2/27 
  b. 7/108 
  c. 14/81 
  d. 1/9

anonymous · Answer

Simple formula: total of *color* / total

kirbykirby · Answer

$$\frac{8}{18}*\frac{7}{18}$$

kirbykirby · Answer

$P(blue)=8/18$ since there are 18 marbles in total, and only 8 are blue
Since you put back the marble, there are again 18 marbles in the bag. Now
$P(red)=7/18$ retricethere are 7 blue marbles out of 18

Now, you want prob(blue) AND prob(red), so you multiply the two probabilities by the product rule

kirbykirby · Answer

With that you should be able to do #2) since it's really the same problem, but just different colours are asked

anonymous · Answer

thank you!

kirbykirby · Answer

:)