Question

a bag contains 5 red marbles, 6 blue marbles and 7 green marbles. if three marbles are drawn out of the bag (without replacement), what is the exact probability that all three marbles drawn will be green?

Explanation:

Step1: Calculate total marbles

The total number of marbles is $5 + 6+7=18$ marbles.

Step2: Calculate first - draw probability

The probability of drawing a green marble on the first draw is $\frac{7}{18}$ since there are 7 green marbles out of 18 total marbles.

Step3: Calculate second - draw probability

After the first green marble is drawn, there are 6 green marbles left and 17 total marbles left. So the probability of drawing a green marble on the second draw is $\frac{6}{17}$.

Step4: Calculate third - draw probability

After the second green marble is drawn, there are 5 green marbles left and 16 total marbles left. So the probability of drawing a green marble on the third draw is $\frac{5}{16}$.

Step5: Calculate combined probability

Since these are dependent events, the probability that all three marbles are green is the product of the probabilities of each draw: $\frac{7}{18}\times\frac{6}{17}\times\frac{5}{16}=\frac{7\times6\times5}{18\times17\times16}=\frac{210}{4896}=\frac{35}{816}$.

Answer:

$\frac{35}{816}$