Charley has a bag of color tiles. In Charley's bag there are 15 red tiles, 4 blue tiles, 1 yellow tile, and 30 green tiles. Charley removed one tile from the bag, then a second tile, and a third tile. What is the probability that the tiles came out as 1 red, 1 green, and then 1 blue in that order?

Question

Charley has a bag of color tiles. In Charley's bag there are 15 red tiles, 4 blue tiles, 1 yellow tile, and 30 green tiles. Charley removed one tile from the bag, then a second tile, and a third tile. What is the probability that the tiles came out as 1 red, 1 green, and then 1 blue in that order?

Answer 1

50 tiles in all to start

1 red:
15/50 = 3/10
now 49 tiles
1 green:
30/49
now 48 tiles
now 1 blue
4/48 = 1/12
so
3/10 * 30/49 * 1/12

Answer 2

find the probability of pulling the outcome 2 x in row with replacing the tiles after each pull .p(blue)

Answer 3

To find the probability of drawing 1 red tile, 1 green tile, and 1 blue tile in that specific order, we need to know the total number of tiles in Charley's bag and the number of each type of tile.

Given:
- 15 red tiles
- 4 blue tiles
- 1 yellow tile
- 30 green tiles

The total number of tiles in the bag is:
15 (red) + 4 (blue) + 1 (yellow) + 30 (green) = 50

To calculate the probability of each successive event, we must consider that each draw is conducted without replacement. This means that the number of potential tiles in the bag decreases after each draw.

For the first draw, there are 50 tiles in total, and we want to draw a red tile. So the probability of drawing a red tile on the first draw is:
P(red on 1st draw) = (number of red tiles) / (total number of tiles) = 15 / 50 = 3/10

After the first draw, there are now 49 tiles in total, and since we successfully drew a red tile, there are now 14 red tiles remaining in the bag.

For the second draw, we want to draw a green tile. So the probability of drawing a green tile on the second draw is:
P(green on 2nd draw) = (number of green tiles) / (total number of remaining tiles) = 30 / 49

After the second draw, there are now 48 tiles in total, and since we successfully drew a green tile, there are now 29 green tiles remaining in the bag.

For the third draw, we want to draw a blue tile. So the probability of drawing a blue tile on the third draw is:
P(blue on 3rd draw) = (number of blue tiles) / (total number of remaining tiles) = 4 / 48 = 1 / 12

Since these draws are independent events, the probabilities multiply together:
P(red, green, blue in order) = P(red on 1st draw) * P(green on 2nd draw) * P(blue on 3rd draw)
= (3/10) * (30/49) * (1/12)
= 9/245