Question

PLEASE help answer all 5 questions

Answer 1

1)

We have 2 red marbles and 9 marbles in total, so the probability of drawing a red marble, is the number of red marbles over the total number of marbles.

`p(r) = (2)/(9)`

2)

We have 3 blue marbles and 9 marbles in total, so the probability of drawing a blue marble, is the number of blue marbles over the total number of marbles.

`p(b) = (3)/(9) = (1)/(3)`

3)

Since we put the marbles back, the sample space stays 9 balls,so we have to multiply the probability of drawing a red marble then a blue one. Since order matters (red before blue) we do not account for permutations.

4)

Putting the marbles back in the bag makes this these events independent, as the sample space does not change.

5)

`p(r \: then \: b) = (2)/(9) * (1)/(3) = (2)/(27)`