Two chips are selected from a box containing 6 blue chips, 4 red chips, and 3 green chips. The first chip selected is not replaced before the second is drawn. Find P(red, then green).

What is (4/13)(3/12) ?

To find the probability (P) of drawing a red chip first (R), and then a green chip (G), we need to calculate the probability of each event happening and multiply these probabilities together.

First, let's calculate the probability of drawing a red chip first. There are a total of 6 + 4 + 3 = 13 chips in the box. Since there are 4 red chips, the probability of drawing a red chip on the first draw is 4/13.

After removing one red chip from the box, we have 13 - 1 = 12 chips left. Out of these remaining chips, there are 3 green chips. Therefore, the probability of drawing a green chip on the second draw, given that a red chip was drawn first, is 3/12.

Since the two events (red and green) are independent, meaning the outcome of one does not affect the other, we can multiply these two probabilities together to find the overall probability:

P(R, then G) = P(R) * P(G | R)

P(R, then G) = (4/13) * (3/12)

Simplifying the fraction gives us:

P(R, then G) = (1/13) * (1/4)

P(R, then G) = 1/ 52

So, the probability of drawing a red chip first and then a green chip is 1/52.