Question

what is the probability drawing 3 green marbles in a row without replacement? 0.00879 0.001234 0.1234

Explanation:

Step1: Assume total marbles and green marbles

Let's assume there are $n$ total marbles and $m$ green marbles. The probability of drawing the first - green marble is $P_1=\frac{m}{n}$.

Step2: Probability of second green marble

After drawing one green marble, there are $n - 1$ total marbles and $m-1$ green marbles. So the probability of drawing the second green marble is $P_2=\frac{m - 1}{n - 1}$.

Step3: Probability of third green marble

After drawing two green marbles, there are $n - 2$ total marbles and $m - 2$ green marbles. So the probability of drawing the third green marble is $P_3=\frac{m - 2}{n - 2}$.

Step4: Overall probability

The probability of drawing 3 green marbles in a row without replacement is $P = P_1\times P_2\times P_3=\frac{m(m - 1)(m - 2)}{n(n - 1)(n - 2)}$.

However, since we are not given the number of marbles, we can't calculate the exact value. But if we assume some values for illustration. Let's say there are 10 marbles in total ($n = 10$) and 4 green marbles ($m = 4$).
$P=\frac{4\times3\times2}{10\times9\times8}=\frac{24}{720}=0.0333$.

Since we don't have enough information to calculate the exact value from the problem - statement, we can't verify which of the given options is correct. But if we assume we had all the necessary information, we would follow the above - mentioned steps to calculate the probability.

Answer:

We need more information (number of total marbles and green marbles) to determine which option is correct.