I just found the compliment the same way I found the answer to A. Number of desired outcomes over number of possible outcomes, which was 48/52 (52 cards minus 4 aces), and I got the answer much quicker. Addition and subtraction with fractions is annoying :/
There are 10 tiles in total. The probability of getting a blue tile with all tiles present is 3/10. Alright, the question says you had successfully drawn a blue tile. So that leaves 2 blue tiles left, and 9 tiles total, b/c you didn't replace any tiles. So now the prob. of drawing a blue tile is 2/9. The questions says "a blue tile AND then another blue tile...". Think of AND as multiply and OR as add. So simply mulitply the two terms together, (3/10)*(2/9)=(6/90). There ya go!
Can Anybody Answer This? Please :] A Bag Contains 4 Red Tiles, 3 Blue Tiles,2 Green Tiles & 1 Yellow... What Is The Probability of drawing a blue tile & then another Blue Tile If The First Tile is NOT returned to the bag? If You Can PLEASE PLEASE Mail me With The Work Please :] But In The Most FRIENDLIEST Way..Don't take me wrong.....
7 years ago... but I'll answer anyway. The probability of the first one being blue is 3/10. The probability of the second being blue is 2/9 (2 blue left, and 9 tokens left). The probability of blue-blue is then (3/10) x (2/9) = 6/90 = 1/15 This is an example of conditional probability.
