A bag contains 6 red marbles, 5 blue marbles and 4 green marbles. If three marbles are drawn out of the bag, what is the exact probability that all three marbles drawn will be blue?
There are a total of 15 marbles in the bag. The probability of drawing a blue marble on the first draw is 5/15. After one blue marble is drawn, there are 4 blue marbles left out of a total of 14 marbles. Therefore, the probability of drawing a second blue marble is 4/14. After two blue marbles are drawn, there are 3 blue marbles left out of a total of 13 marbles. Therefore, the probability of drawing a third blue marble is 3/13.
To find the probability of drawing all three marbles as blue, we need to multiply the probabilities of each individual draw:
P(all three marbles drawn are blue) = (5/15) x (4/14) x (3/13)
Simplifying this expression, we get:
P(all three marbles drawn are blue) = 1/91
Therefore, the exact probability that all three marbles drawn will be blue is 1/91.
To find the probability of drawing three blue marbles, we need to consider the total number of marbles and the number of favorable outcomes.
The total number of marbles in the bag is 6 (red) + 5 (blue) + 4 (green) = 15 marbles.
To calculate the probability of drawing three blue marbles, we need to consider the number of ways to select 3 blue marbles out of the total number of marbles.
The number of favorable outcomes is choosing 3 blue marbles out of the 5 blue marbles, which can be calculated using the combination formula:
C(5, 3) = 5! / (3! * (5-3)!) = 10
The total number of possible outcomes is choosing any 3 marbles out of the 15 marbles in the bag:
C(15, 3) = 15! / (3! * (15-3)!) = 455
Therefore, the probability of drawing three blue marbles is:
P(3 blue marbles) = favorable outcomes / total outcomes = 10 / 455 = 2 / 91
So, the exact probability of drawing three blue marbles is 2/91.