A bag contains 4 red marbles, 3 blue marbles, and 6 green marbles. If three marbles are drawn out of the bag, what is the exact probability that all three marbles drawn will be red... A bag contains 4 red marbles, 3 blue marbles, and 6 green marbles. If three marbles are drawn out of the bag, what is the exact probability that all three marbles drawn will be red? Read more

Steps to Solve

1. Identify the total number of marbles

First, calculate the total number of marbles in the bag.

The bag contains:

• 4 red marbles
• 3 blue marbles
• 6 green marbles

So, the total number of marbles is: $$ 4 + 3 + 6 = 13 $$

2. Calculate the total ways to choose 3 marbles

Next, determine how many ways we can choose 3 marbles from 13. This can be found using combinations:

$$ C(n, r) = \frac{n!}{r!(n-r)!} $$

Where ( n ) is the total number of marbles (13) and ( r ) is the number of marbles chosen (3).

Calculating:

$$ C(13, 3) = \frac{13!}{3!(13-3)!} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286 $$

3. Calculate the ways to choose 3 red marbles

Now, calculate how many ways to choose 3 red marbles from the 4 available:

$$ C(4, 3) = \frac{4!}{3!(4-3)!} = \frac{4}{1} = 4 $$

4. Calculate the probability

Finally, find the probability that all three drawn marbles will be red. The probability ( P ) is given by the formula:

$$ P(\text{all red}) = \frac{\text{Ways to choose 3 red}}{\text{Total ways to choose 3 marbles}} = \frac{C(4, 3)}{C(13, 3)} $$

Substituting the values calculated:

$$ P(\text{all red}) = \frac{4}{286} = \frac{1}{71.5} \approx 0.014 $$