Peter purchased a pentagonal pen for his puppy Piper. Now Peter wants to decorate the new pen for Piper, and he would like to paint each side of the pen either red, green, or blue so that each wall is a solid color.
Peter can only paint at night when Piper is sleeping, and unfortunately, it is too dark for him to determine which color he is painting. So for each wall, Peter randomly chooses a can of paint and paints the wall in that color. In the morning, Peter observes the resulting color scheme. The vertices of the pentagon are labeled with the letters and, and these labels are clearly visible during the daytime. What is the probability that no two adjacent walls of the pen have the same color?
I got 243 total ways with no restrictions, and 48 possibilities counting adjacent. so I got 16/81. Can someone please give me a detailed answer that's understandable.
Lets do it this way
Find the total number of ways you can paint the wall.
Then find the number of favorable cases over the total cases.
So lets have walls
1 (3 ways to paint it red, green or blue)
2 (3 ways)
3 (3 ways)
4 (3 ways)
5 (3 ways)
So 3^5 = 243 total ways you can paint it
Now we try to find the number of ways we can paint it that has no two adjacent walls that are same color
1 (3 ways to paint it red, green, or blue)
2 (2 ways to paint it, different than first one)
3 (2 ways to paint it, different than second one)
4 (2 ways to paint it, different than third one)
5 ( 1 way to paint it different than first and fourth)
So 3 * 2 * 2 * 2 * 1 = 24
HOWEVER, the above example is when the first wall is DIFFERENT than the fourth wall.
So we also have to add cases where they are the same.
1 (3 ways to paint it red, green, or blue)
2 (2 ways to paint it, different than first one)
3 (2 ways to paint it, different than second one)
4 (1 ways to paint it, same as first)
5 ( 2 ways to paint it, because the first and fourth are the same)
So 3 * 2 * 2 * 1 * 2 = 24
So 24 + 24 = 48 ways to paint it with no adjacent wall same
48/243 = 16/81
So the probability is 16/81.