Please help! 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 A, B, C, D, and E, 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?

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Ritz01 Ritz01 · Answer 1 · 2020-10-27T12:52:56+01:00

Answer:

16/81 = 0.20 (approximately)

Step-by-step explanation:

Denote the wall AB by 1, wall BC by 2, wall CD by 3, wall DE by 4 and wall EA by 5.

Wall 1 can be painted in 3 ways i.e. either by Red or by Green or by Blue.

Similarly, wall 2, 3, 4 and 5 can be painted by 3 ways.

So, the total number of possible case [tex]= 3^5=243[/tex].

Now, for no two adjecent walls to have the same color, total number of favourable cases are

Wall 1 can be painted in 3 ways (Red or Green or Blue).

Wall 2 can be painted in 2 ways (except the color of wall 1),

Wall 3 can be painted in 2 ways (except the color of wall 2),

Wall 4 can be painted in 2 ways (except the color of wall 3),

Wall 5 can be painted in 1 ways (if the color of walls 1 and 4 are the different)

For this case, the number of favourable cases = 3x2x2x2x1=24.

If the color of the walla 1 and 4 are the same, then the wall 3 can be painted in 2 way to have different color from adjecent walls, i.e

Wall 1 can be painted in 3 ways (Red or Green or Blue).

Wall 2 can be painted in 2 ways (except the color of wall 1),

Wall 3 can be painted in 1 ways (except the color of walls 1 and 4 which have different color),

Wall 4 can be painted in 2 ways (except the color of wall 3),

Wall 5 can be painted in 2 ways (as the color of walls 1 and 4 are the different)

For this case, the number of favourable cases = 3x2x1x2x2=24.

So, the total number of favourable cases= 24+24=48.

As the probability of favourable event = (Total number of possible events)/(Total number of favourable cases)

=48/343

=16/81

Hence, the probability that no two adjacent walls of the pen have the same color is 16/81 = 0.20 (approximately).