Counting Dependent Events
Reference > Mathematics > Algebra > Counting PrinciplesIn the previous section, we looked at how to count independent events using the fundamental counting principle. In this section we'll explore what happens when the events are not independent.
If two events are not independent, that means that what happens in one event affects what happens in the next. If this is the case, we call them dependent events.
Suppose I was painting my kitchen, selecting new cabinets, and putting down floor tile. I have 5 color paints to choose from, 3 styles of cabinets, and 4 patterns of floor tiles. However, as I look at the items together, I decide that the green wall paint doesn't go well with the blue floor tiles. In how many ways can I design my kitchen?
If I hadn't decided that I can't mix green and blue, I'd have 5 x 3 x 4 = 60 ways of designing my kitchen. But my color condition throws a monkey wrench in and makes it a bit more complicated.
So we'll break it up into two cases. In the first case, I'm going to pick green for my wall color, and in the second case, I'm going to pick any other color except green.
Case #1
1 way to pick the wall color, 3 cabinet choices, and 3 floor tile choices (because I left out the blue one). 1 x 3 x 3 = 9
Case #2
4 ways to pick the wall color (because I left out the green), 3 cabinet choices, and 4 floor tile choices. 4 x 3 x 4 = 48
Now we combine them: 48 + 9 = 57.
57 makes perfect sense; there were 3 choices we eliminated (any one of the three cabinet styles with the green/blue combination). If you add those 3 back in, you get 60, which is what we calculated for independent events.
Here's another example of dependent events. Suppose you have a hat, and you put ten slips of paper in it, each with a number from 0 to 9 on it. Now you draw three slips from the hat (we call this "drawing withour replacement," because we're not putting the slip back in after we draw it from the hat). In how many ways can you do that?
There are ten ways to pull the first slip, but then there are only nine left, which means there are only nine ways to pull the second slip. Similarly, there are eight ways to pull the third slip. Thus, there are 10 x 9 x 8 = 720 ways to pull three slips from the hat.