Question

Suppose three fair dice are rolled. What is the probability at most one six appears?

Short answer

The probability that at most one six appears when three fair dice are rolled is \(\frac{25}{27}\).

Step-by-step solution

Define events and complementary events

Let A be the event "at most one six appears." We are asked to find P(A). Let A' be the event "at least two sixes appear" (the complementary event). Then, we can express A as the complement of A': P(A) = 1 - P(A').

Calculate the probability of each outcome for a single dié

Since each die is fair and has six sides, the probability of rolling a six is 1/6 and the probability of not rolling a six is 5/6.

Compute the probability of A' using combinations

A' can occur when either two or three sixes appear. We can calculate the probability of these situations and sum them: a) Two sixes and one non-six: There are \(\binom{3}{2}\) = 3 ways to choose which two dice will have the sixes. The probability of this configuration is \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\). b) All three sixes: There is only one way that all three dice can have sixes, which has probability \(\left(\frac{1}{6}\right)^3\).

Compute P(A')

Now let's sum the probabilities from Step 3 to find P(A'): P(A') = 3 * \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\) + \(\left(\frac{1}{6}\right)^3\).

Compute P(A) using the complement

Now that we have P(A'), we can compute P(A) using the complement relationship: P(A) = 1 - P(A') = 1 - [3 * \(\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)\) + \(\left(\frac{1}{6}\right)^3\)].

Evaluate and simplify

Evaluate and simplify the expression from Step 5: P(A) = 1 - [3 * \(\left(\frac{1}{36}\right)\left(\frac{5}{6}\right)\) + \(\frac{1}{216}\)] = 1 - [\(\frac{5}{72}\) + \(\frac{1}{216}\)]. First, convert the fractions to a common denominator (the least common multiple of 72 and 216 is 216): P(A) = 1 - [\(\frac{15}{216}\) + \(\frac{1}{216}\)]. Now, we can simplify the expression by combining the fractions: P(A) = 1 - \(\frac{16}{216}\). Simplify further by dividing by the greatest common divisor (8): P(A) = 1 - \(\frac{2}{27}\). Finally, convert the mixed number into an improper fraction: P(A) = \(\frac{27 - 2}{27}\) = \(\frac{25}{27}\).

State the final answer

So, the probability that at most one six appears when three fair dice are rolled is \(\frac{25}{27}\).