Final answer:
The probability of the sum of the numbers on the bottom faces of two tetrahedrons being 6 is 1/16 or 0.0625. The most likely sum is 3, with a probability of 4/16 or 0.25.
Step-by-step explanation:
The student's question involves finding probabilities related to rolling two regular tetrahedrons each marked with the numbers 0, 1, 2, and 3 on its faces. Let's address each part of the question step by step:
Part (a): Probability that the sum is 6
To find the probability of the sum being 6, we must consider all possible outcomes when two tetrahedrons are rolled. The total number of outcomes when rolling two tetrahedrons is 4 times 4, which equals 16, since there are 4 faces on each tetrahedron. However, since the tetrahedrons have numbers 0 through 3, the highest sum we can get is 3 + 3 = 6. The only way to get a sum of 6 is if both tetrahedrons land on their faces with the number 3. There is only 1 such favorable outcome. Therefore, the probability of the sum being 6 is 1/16 or 0.0625.
Part (b): Most Likely Sum and Its Probability
The most likely sum when rolling two tetrahedrons would be the middle values, as there are multiple ways to achieve these sums. The sums in the middle of the possible range (0 to 6) are 2 and 3. By calculating the probabilities for each sum, we find that the sum of 3 has the highest number of favorable outcomes (0+3, 1+2, 2+1, 3+0), which is 4. The probability for the most likely sum, which is 3, is therefore 4/16 or 0.25.