We roll a fair 6-sided die 5 times. What is the probability that we get a 6 in at most 2 of the rolls?

Amanita, let's break down the problem step by step:

Step 1: Understand the problem

We roll a fair 6-sided die 5 times. We want the probability of getting a 6 in at most 2 of those rolls. "At most 2" means we can get a 6 zero times, one time, or two times across the 5 rolls.

Step 2: Choose the proper probability model

This scenario can be described using the binomial probability formula, because each roll is independent and there is only "success" (getting a 6) or "failure" (getting anything else) in each roll.

The binomial probability formula is:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

where:

  • nn = total number of trials (rolls) = 5
  • kk = number of desired successes (times we get a 6) = 0, 1, or 2
  • pp = probability of success on a single roll = 16\frac{1}{6}

Step 3: Write the probability expression for "at most 2 sixes"

We need:

P(X2)=P(X=0)+P(X=1)+P(X=2)P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

Step 4: Calculate each term using the binomial formula

Let's substitute the values for each case:

  • For k=0k=0: P(X=0)=(50)(16)0(56)5P(X=0) = \binom{5}{0} (\frac{1}{6})^0 (\frac{5}{6})^5
  • For k=1k=1: P(X=1)=(51)(16)1(56)4P(X=1) = \binom{5}{1} (\frac{1}{6})^1 (\frac{5}{6})^4
  • For k=2k=2: P(X=2)=(52)(16)2(56)3P(X=2) = \binom{5}{2} (\frac{1}{6})^2 (\frac{5}{6})^3

Step 5: Sum all the probabilities to find the answer

Now, I will calculate and sum the values for P(X=0)P(X=0), P(X=1)P(X=1), and P(X=2)P(X=2). Amanita, here is how we solve the problem step by step:

  1. We use the binomial probability formula to find the probability of getting a 6 in at most 2 rolls when we roll a die 5 times.
  2. We calculate the probability for getting 0, 1, or 2 sixes, and add them up.
  3. The formula used is:

P(X=k)=(5k)(16)k(56)5k P(X = k) = \binom{5}{k} \left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{5-k}

for k=0,1,2k = 0, 1, 2. 4. Adding these probabilities, we get:

P(X2)=P(X=0)+P(X=1)+P(X=2)=31257776+31257776+6253888=625648 P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = \frac{3125}{7776} + \frac{3125}{7776} + \frac{625}{3888} = \frac{625}{648}

Final Answer:
The probability that you get a 6 in at most 2 of the 5 rolls is

625648\boxed{\frac{625}{648}}