To calculate the probability of rolling exactly two 1's when rolling six six-sided dice, we can use the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

where:

• $n$ is the total number of trials (dice rolled),
• $k$ is the number of successful outcomes (rolling a 1),
• $p$ is the probability of success on a single trial (rolling a 1),
• $\binom{n}{k}$ is the binomial coefficient.

1. Parameters:

   • $n = 6$ (the number of dice),
   • $k = 2$ (the number of 1's we want to roll),
   • $p = \frac{1}{6}$ (the probability of rolling a 1 on one die).

2. Calculate the binomial coefficient: $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$

3. Calculate the probabilities: $p^k = \left( \frac{1}{6} \right)^2 = \frac{1}{36}$ $(1 - p)^{n - k} = \left( \frac{5}{6} \right)^{4} = \left( \frac{625}{1296} \right)$

4. Combine these values: $P(X = 2) = \binom{6}{2} \times \left( \frac{1}{6} \right)^2 \times \left( \frac{5}{6} \right)^4$ $P(X = 2) = 15 \times \frac{1}{36} \times \frac{625}{1296}$ $= \frac{15 \times 625}{36 \times 1296}$ $= \frac{9375}{46656}$ $\approx 0.2004$

Thus, the probability of rolling exactly two 1's when rolling six six-sided dice is approximately 0.2004, or 20.04%.

Would you like more details or have any questions? Here are some related questions you might consider:

1. What is the probability of rolling three 1's?
2. How does the probability change with a different number of dice?
3. What is the expected number of 1's when rolling six dice?
4. How would you calculate the probability of rolling no 1's?
5. Can you explain the binomial distribution further?

Tip: When calculating probabilities for multiple events, always consider the total number of possible outcomes.