Multiple choice

Three six-faced fair dice are thrown together. The probability that the sum of the numbers appearing on the dice is K $(3\leq K \leq8)$ is

  1. $\displaystyle \frac { (k-1) (k-2) }{432 } $
  2. $\displaystyle \frac{k(k-1)}{432}$
  3. $\displaystyle \frac{k^2}{432}$

  4. None of these
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A Correct answer
Explanation

The number of ways to get a sum K with three dice is the coefficient of x^K in (x + x^2 + x^3 + x^4 + x^5 + x^6)^3. For 3 <= K <= 8, this is given by the formula (K-1)(K-2)/2. Dividing by the total outcomes 6^3 = 216 gives (K-1)(K-2)/432.

AI explanation

The total number of outcomes when three dice are thrown is 6^3, which equals 216. The number of favorable ways to obtain a sum K, where K is between 3 and 8, is given by the combinatorial identity for integer solutions as (K-1)(K-2)/2. Therefore, the required probability is ((K-1)(K-2)/2) divided by 216, which simplifies to (K-1)(K-2)/432.

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