A bag contains 6 red marbles, 8 blue marbles and 4 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 red?

1 answer

To find the probability that all three marbles drawn from the bag are red, we first need to calculate the total number of marbles in the bag. The bag contains:

  • 6 red marbles
  • 8 blue marbles
  • 4 green marbles

Calculating the total number of marbles:

\[ \text{Total marbles} = 6 + 8 + 4 = 18 \]

Next, we will calculate the probability of drawing 3 red marbles out of 6 red marbles. The number of ways to choose 3 red marbles from 6 is given by the combination formula \( \binom{n}{k} \), which represents the number of ways to choose \( k \) objects from \( n \) objects without regard to the order of selection.

\[ \text{Ways to choose 3 red marbles from 6} = \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

Next, we need to calculate the total number of ways to choose any 3 marbles from 18 marbles:

\[ \text{Ways to choose 3 marbles from 18} = \binom{18}{3} = \frac{18!}{3!(18-3)!} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816 \]

Now, the probability that all three marbles drawn will be red is the ratio of the number of favorable outcomes (choosing 3 red marbles) to the total outcomes (choosing any 3 marbles):

\[ P(\text{3 red marbles}) = \frac{\text{Ways to choose 3 red marbles}}{\text{Ways to choose 3 marbles from 18}} = \frac{20}{816} \]

To simplify this fraction, we can divide both the numerator and the denominator by 4:

\[ \frac{20 \div 4}{816 \div 4} = \frac{5}{204} \]

Thus, the exact probability that all three marbles drawn will be red is:

\[ \boxed{\frac{5}{204}} \]