Question

A jar has 19 marbles and 7 of them are green. You pull 3 marbles without replacement. What is the probability that all are green? Write your answer as a fraction. You do not need to simplify.

Answer 1

To find the probability that all 3 marbles drawn are green, we will use the formula for probabilities in drawing without replacement.

1. **Total number of marbles**: 19
2. **Number of green marbles**: 7

We need to calculate the probability of drawing 3 green marbles in a row without replacement.

- The probability of drawing the first green marble:
\[
P(\text{1st green}) = \frac{7}{19}
\]

- After drawing one green marble, there are now 6 green marbles left and a total of 18 marbles:
\[
P(\text{2nd green | 1st green}) = \frac{6}{18} = \frac{1}{3}
\]

- After drawing two green marbles, there are now 5 green marbles left and a total of 17 marbles:
\[
P(\text{3rd green | 1st and 2nd green}) = \frac{5}{17}
\]

Now, we multiply these probabilities together to find the overall probability of drawing 3 green marbles:
\[
P(\text{all 3 green}) = P(\text{1st green}) \times P(\text{2nd green | 1st green}) \times P(\text{3rd green | 1st and 2nd green})
\]

Calculating this, we have:
\[
P(\text{all 3 green}) = \frac{7}{19} \times \frac{6}{18} \times \frac{5}{17}
\]

Now, we multiply the numerators and the denominators:
\[
P(\text{all 3 green}) = \frac{7 \times 6 \times 5}{19 \times 18 \times 17}
\]

Calculating the numerator:
\[
7 \times 6 = 42,
\]
\[
42 \times 5 = 210
\]

Calculating the denominator:
\[
19 \times 18 = 342,
\]
\[
342 \times 17 = 5814
\]

Thus, the probability that all three marbles drawn are green is:
\[
\frac{210}{5814}
\]

The final answer is:
\[
\frac{210}{5814}
\]