Question

A bag contains 3 red marbles, 6 blue marbles and 4 green marbles. If three marbles are drawn out of the bag (without replacement), what is the probability, to the nearest 1000th, that all three marbles drawn will be blue?

Answer 1

To find the probability that all three marbles drawn from the bag are blue, we start by determining the total number of marbles in the bag.

The bag contains:
- 3 red marbles
- 6 blue marbles
- 4 green marbles

The total number of marbles is:

\[
3 + 6 + 4 = 13
\]

Next, we need to calculate the total number of ways to choose 3 marbles from the 13 marbles, which can be calculated using the combination formula:

\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\]

where \( n \) is the total number of items, and \( k \) is the number of items to choose.

The total ways to choose 3 marbles from 13 is:

\[
\binom{13}{3} = \frac{13!}{3!(13-3)!} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286
\]

Now, we need to find the number of ways to choose 3 blue marbles from the 6 available blue marbles. The number of ways to choose 3 blue marbles is:

\[
\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20
\]

Now, to find the probability of drawing 3 blue marbles, we can use the formula for probability:

\[
P(\text{all blue}) = \frac{\text{Number of ways to choose 3 blue}}{\text{Total ways to choose 3 marbles}} = \frac{\binom{6}{3}}{\binom{13}{3}} = \frac{20}{286}
\]

We can simplify this fraction:

\[
\frac{20}{286} = \frac{10}{143}
\]

Next, we convert this fraction to a decimal for calculating the probability:

\[
\frac{10}{143} \approx 0.0699301
\]

Finally, to express this probability to the nearest thousandth, we round 0.0699301:

\[
0.0699301 \approx 0.070
\]

Therefore, the probability that all three marbles drawn will be blue is:

\[
\boxed{0.070}
\]