Question

10. A bag contains 8 red marbles and 12 blue marbles. One marble is chosen at random and returned to the bag. Then, a second marble is chosen at random.

What is the probability that the first marble chosen is blue and the second marble chosen is red?

Answer 1

0.6&0.4

Answer 2

Both are occurring one after the other. So we have to consider the draw as part of the same event.

So for drawing the first marble

total no. of marbles = 20
No. of red = 8
No. of blue = 20

So first one being blue P (Blue) = 12/20 = 3/5

Now the marble is replaced so the total does not change.

Second one being red P(Red) = 8/20 = 2/5

So the combined probability is = 3/5 times 2/5 = 6/25

Answer 3

If you haven't wrapped your head around @rajee_sam's fine explanation (my only quibble is that he wrote "No. of blue = 20" where he meant "No. of blue = 12"), here's a different way to think about it.

We choose 1 of 20 marbles twice in a row, so there are 20*20 = 400 different outcomes. We need to count all the possible successful outcomes, then divide by the total number of outcomes, and that will give us the probability.

If the first marble we draw is red (8 out of 20 cases), we've failed. Each of those outcomes has 20 unsuccessful outcomes (no matter what color we draw for the second marble), so that accounts for 8*20 = 160 outcomes.

If the first marble is blue (12 out of 20 cases), then the second marble we draw is of interest. For each of those 12 cases, we'll have 8 red 2nd marbles and 12 blue 2nd marbles. The red 2nd marbles are successful outcomes (12 * 8 = 96 successful outcomes) and the blue 2nd marbles are not (12 * 12 = 144 unsuccessful outcomes).

In all, we have 96 successful outcomes out of 400 possible outcomes. Our probability of success is thus 96/400 or 6/25.