Question

A jar contains 4 red marbles, 4 green marbles, and 5 blue marbles. If we choose a marble, then another marble without putting the first one back in the jar, what is the probability that the first marble will be blue and the second will be green?

• A. $\frac{4}{39}$
• B. $\frac{5}{39}$
• C. $\frac{6}{37}$
• D. $\frac{3}{37}$

Answer 1

The correct option is B $\frac{5}{39}$

The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A has already happened.
In this case, event A is picking a blue marble and leaving it out. Event B is picking a green marble.
Let's take the events one at a time. What is the probability that the first marble chosen will be blue?
There are 5 blue marbles, and 13 total, so the probability we will pick a blue marble is $\frac{5}{13}$.
After we take out the first marble, we don't put it back in, so there are only 12 marbles left.
Since the first marble was blue, there are still 4 green marbles left.
So, the probability of picking a green marble after taking out a blue marble is $\frac{4}{12}$
Therefore, the probability of picking a blue marble, then a green marble is $\frac{5}{13}\ast \frac{4}{12}=\frac{20}{156}=\frac{5}{39}$