+34 Each face of a cube is painted randomly one of the colors red, orange, yellow, green, blue or purple. What is the probability that the cube has at least one pair of faces that share an edge and are the same color? Express your answer as a decimal to the nearest thousandth.
+42 Let's solve this problem step by step.
1. Total Number of Ways to Paint the Cube
Each of the 6 faces can be painted in 6 different colors.
Total number of ways to paint the cube is 6^6 = 46656.
2. Finding the Complement: Probability of No Adjacent Faces with the Same Color
It's easier to find the probability of the complement (no adjacent faces have the same color) and subtract it from 1.
First Face: We can choose any of the 6 colors.
Opposite Face: We can choose any of the 6 colors.
Remaining 4 Faces:
The remaining 4 faces form a ring.
The first of these 4 faces can be any of the 5 remaining colors (not the color of the adjacent face).
The second of these 4 faces can be any of the 5 remaining colors (not the color of the adjacent face).
The third of these 4 faces can be any of the 5 remaining colors (not the color of the adjacent face).
The fourth of these 4 faces can be any of the 5 remaining colors (not the color of the adjacent face).
However, we have to consider the first and last of these 4 faces. If they are the same color, the color of the 3rd face can be any of 5 colors. If they are different, the color of the 3rd face can be any of 4 colors.
Case 1: All 4 faces in the ring have different colors.
The first face can be any of 5 colors.
The second face can be any of 4 colors.
The third face can be any of 3 colors.
The fourth face can be any of 2 colors.
Total: 5 * 4 * 3 * 2 = 120
Case 2: The 1st and 3rd faces have the same color, and the 2nd and 4th faces have the same color.
The 1st face can be any of 5 colors.
The 2nd face can be any of 4 colors.
The 3rd face must be the same as the 1st (1 choice).
The 4th face must be the same as the 2nd (1 choice).
Total: 5 * 4 * 1 * 1 = 20
Case 3: The 1st and 3rd faces have the same color, but the 2nd and 4th faces are different.
The 1st face can be any of 5 colors.
The 2nd face can be any of 4 colors.
The 3rd face must be the same as the 1st (1 choice).
The 4th face can be any of 3 colors (different from 1st and 2nd).
Total: 5 * 4 * 1 * 3 = 60
Case 4: The 2nd and 4th faces have the same color, but the 1st and 3rd faces are different.
The 1st face can be any of 5 colors.
The 2nd face can be any of 4 colors.
The 3rd face can be any of 3 colors (different from 1st and 2nd).
The 4th face must be the same as the 2nd (1 choice).
Total: 5 * 4 * 3 * 1 = 60
Case 5: The 1st and 4th faces have the same color, but the 2nd and 3rd faces are different.
The 1st face can be any of 5 colors.
The 2nd face can be any of 4 colors.
The 3rd face can be any of 3 colors (different from 1st and 2nd).
The 4th face must be the same as the 1st (1 choice).
Total: 5 * 4 * 3 * 1 = 60
Case 6: The 2nd and 3rd faces have the same color, but the 1st and 4th faces are different.
The 1st face can be any of 5 colors.
The 2nd face can be any of 4 colors.
The 3rd face must be the same as the 2nd (1 choice).
The 4th face can be any of 3 colors (different from 1st and 2nd).
Total: 5 * 4 * 1 * 3 = 60
Case 7: Two pairs of adjacent faces are the same color.
The first face can be any of 5 colors.
The adjacent face can be any 4 colors.
The remaining two faces can be any 1 color.
541 = 20. But order matters, so 5 * 4 = 20
Case 8: All four faces have different colors.
543*2 = 120
Case 9: The first and third faces have the same color, and the second and fourth faces have the same color.
5 * 4 = 20
Case 10: The first and fourth faces have the same color, and the second and third faces have the same color.
5 * 4 = 20
Total number of ways: 6 * 6 * (120 + 20 + 60 + 60) = 36 * 260 = 9360
Total Ways: 6 * 6 * (5432 + 543 + 54) = 36*200
Number of ways = 65432 = 720.
Number of ways to color the ring = (5432) + (5411) = 120+20 = 140.
Ways to paint the cube: 66140 = 5040
Probability of No Adjacent Faces with Same Color: 5040/46656 = 0.108
3. Probability of At Least One Pair of Adjacent Faces with Same Color
P(At least one pair) = 1 - P(No adjacent pairs)
P(At least one pair) = 1 - 5040/46656 ≈ 1 - 0.1080 = 0.892
Therefore, the probability is approximately 0.892.
+34 Thank you for trying, but you're answer is not correct. The correct answer was 0.913.
We can use complementary counting to solve this problem, by subtracting the probabilty that no adjacent faces are the same color:
1: All faces are colored diffferently. 6!=720 ways
2: One pair of opposite faces are colored the same. 3 ways to choose the pair of faces * 6 ways to color the faces * 120 ways to color the remaining faces = 2160 ways.
3: Two pairs of opposite faces are colored the same. 3 ways to choose the pairs of faces * 30 ways to color the faces * 12 ways to color the remaining faces = 1080 ways.
4: Each pair of opposite faces are the same. 6 * 5 * 4 ways to choose the colors = 120 ways.
(720 + 2160 + 1080 + 120) / 6^6 (total number of ways) ~ 0.08744.
1 - 0.08744 ~ 0.913.
Thank you for trying, and I hope this helped!
