A die is thrown 4 times. Find the probability of getting at most two 6.

Explanation

To solve the problem of finding the probability of getting at most two 6s when a die is thrown 4 times, we can follow these steps:

### Step 1: Understand the Problem
We need to find the probability of getting at most two 6s when a die is thrown 4 times. This means we want to calculate the probabilities of getting 0, 1, or 2 sixes.

### Step 2: Define the Probability of Getting a 6
The probability of rolling a 6 on a single die is:
\[ P(6) = \frac{1}{6} \]
The probability of not rolling a 6 is:
\[ P(\text{not 6}) = 1 - P(6) = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 3: Calculate the Probability of Getting 0 Sixes
The probability of getting 0 sixes in 4 throws is:
\[ P(X = 0) = \left(\frac{5}{6}\right)^4 \]

### Step 4: Calculate the Probability of Getting 1 Six
The probability of getting exactly 1 six in 4 throws can be calculated using the binomial probability formula:
\[ P(X = 1) = \binom{4}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^{3} \]
Where \( \binom{4}{1} \) is the number of ways to choose which throw will be a 6.

### Step 5: Calculate the Probability of Getting 2 Sixes
Similarly, the probability of getting exactly 2 sixes is:
\[ P(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{2} \]

### Step 6: Combine the Probabilities
Now, we add the probabilities of getting 0, 1, and 2 sixes:
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]

### Step 7: Calculate Each Probability
1. Calculate \( P(X = 0) \):
\[ P(X = 0) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296} \]

2. Calculate \( P(X = 1) \):
\[ P(X = 1) = \binom{4}{1} \left(\frac{1}{6}\right)^1 \left(\frac{5}{6}\right)^{3} = 4 \cdot \frac{1}{6} \cdot \left(\frac{125}{216}\right) = \frac{500}{1296} \]

3. Calculate \( P(X = 2) \):
\[ P(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^{2} = 6 \cdot \frac{1}{36} \cdot \left(\frac{25}{36}\right) = \frac{150}{1296} \]

### Step 8: Sum the Probabilities
Now, we sum these probabilities:
\[ P(X \leq 2) = \frac{625}{1296} + \frac{500}{1296} + \frac{150}{1296} = \frac{1275}{1296} \]

### Step 9: Final Probability
Thus, the probability of getting at most two 6s when a die is thrown 4 times is:
\[ P(X \leq 2) = \frac{1275}{1296} \approx 0.984 \]