Multiple choice

Three six faced dice are thrown together. The probability that sum of the numbers appearing on the dice is equal to $ \displaystyle k\left ( 3\leq k\leq 4\right ) $ is

  1. $ \displaystyle \frac{k^{2}}{108} $
  2. $ \displaystyle \frac{k\left ( k-1 \right )}{216} $
  3. $ \displaystyle \frac{\left ( k-1 \right )\left ( k-2 \right )}{2\times 6^{3}} $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For three dice, the total outcomes are 6^3 = 216. The number of ways to get a sum k is the coefficient of x^k in (x + x^2 + x^3 + x^4 + x^5 + x^6)^3. For k=3, ways = 1. For k=4, ways = 3. The formula (k-1)(k-2)/2 gives 1 for k=3 and 3 for k=4. Thus, the probability is (k-1)(k-2) / (2 * 6^3).

AI explanation

When three dice are thrown, the number of ways to get a sum of k (where 3 <= k <= 4) is found by distributing (k - 3) among the three dice, giving (k - 1)(k - 2) / 2 combinations. Since the total number of outcomes is 6^3 = 216, the probability is (k - 1)(k - 2) / (2 * 6^3). The probability is (k - 1)(k - 2) / (2 * 6^3).

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