A loaded dice has following probability distribution of occurrences Dice Value 1 2 3 4 5 6 Probability $$\frac{1}{4}$$ $$\frac{1}{8}$$ $$\frac{1}{8}$$ $$\frac{1}{8}$$ $$\frac{1}{8}$$ $$\frac{1}{4}$$ If three identical dice as the above are thrown, the probability of occurrence of values 1, 5 and 6 on the three dice is A. same as that of occurrence of 3, 4, 5 B. same as that of occurrence of 1, 2, 5 C. $$\frac{1}{{128}}$$ D. $$\frac{5}{8}$$

The correct answer is $\boxed{\frac{1}{128}}$.

The probability of occurrence of a particular outcome in a probability distribution is the product of the probabilities of each of the events that must occur for the outcome to occur. In this case, the outcome is getting 1, 5, and 6 on three identical dice. The events that must occur for this outcome to occur are:

• The first die must come up 1.
• The second die must come up 5.
• The third die must come up 6.

The probability of the first die coming up 1 is $\frac{1}{4}$. The probability of the second die coming up 5 is $\frac{1}{8}$. The probability of the third die coming up 6 is $\frac{1}{4}$. Therefore, the probability of getting 1, 5, and 6 on three identical dice is $\frac{1}{4} \times \frac{1}{8} \times \frac{1}{4} = \frac{1}{128}$.

Option A is incorrect because the probability of getting 3, 4, and 5 on three identical dice is $\frac{1}{8} \times \frac{1}{8} \times \frac{1}{4} = \frac{1}{256}$.

Option B is incorrect because the probability of getting 1, 2, and 5 on three identical dice is $\frac{1}{4} \times \frac{1}{8} \times \frac{1}{4} = \frac{1}{64}$.

Option D is incorrect because the probability of getting 5 on three identical dice is $\frac{1}{8} \times \frac{1}{8} \times \frac{1}{4} = \frac{1}{256}$.