Question

two marbles are drawn randomly. what is the probability of drawing a green, followed by a green, if the first one is not replaced before the second? assume the first marble drawn is green. draw #1 draw #2 green marbles: ? / 12 green marbles: / 11 total marbles: total marbles:

Explanation:

Step1: Count green marbles initially

There are 5 green marbles out of 12 total marbles for the first - draw.
So the probability of drawing a green marble on the first draw is $\frac{5}{12}$.

Step2: Calculate probability for second draw

Since the first - drawn green marble is not replaced, there are now 4 green marbles left out of 11 total marbles for the second draw. The probability of drawing a green marble on the second draw, given that the first one was green, is $\frac{4}{11}$.

Step3: Use multiplication rule for dependent events

The probability of both events (drawing a green marble first and then a green marble second without replacement) is the product of the probabilities of each event. So we multiply $\frac{5}{12}$ and $\frac{4}{11}$.
$\frac{5}{12}\times\frac{4}{11}=\frac{5\times4}{12\times11}=\frac{20}{132}=\frac{5}{33}$

Answer:

$\frac{5}{33}$