To calculate the probability \( P(X \leq 1) \) for a binomial experiment where \( n = 12 \) and \( p = 0.3 \), we need to find the probabilities for \( X = 0 \) and \( X = 1 \).
The binomial probability formula is given by:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( \binom{n}{k} \) is the binomial coefficient, which can be calculated as:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Step 1: Calculate \( P(X = 0) \)
Substituting \( k = 0 \): \[ P(X = 0) = \binom{12}{0} (0.3)^0 (0.7)^{12} \] \[ = 1 \cdot 1 \cdot (0.7)^{12} \] \[ = (0.7)^{12} \] \[ \approx 0.1934917633 \quad (\text{using a calculator to approximate} (0.7)^{12}) \]
Step 2: Calculate \( P(X = 1) \)
Substituting \( k = 1 \): \[ P(X = 1) = \binom{12}{1} (0.3)^1 (0.7)^{11} \] \[ = 12 \cdot (0.3) \cdot (0.7)^{11} \] \[ = 12 \cdot 0.3 \cdot (0.7)^{11} \] \[ = 3.6 \cdot (0.7)^{11} \] \[ \approx 3.6 \cdot 0.279936 = 1.0087936 \quad (\text{using a calculator for} (0.7)^{11}) \]
Step 3: Combine \( P(X = 0) \) and \( P(X = 1) \)
Now we find \( P(X \leq 1) = P(X = 0) + P(X = 1) \): \[ P(X \leq 1) = (0.7)^{12} + 12 \cdot (0.3) \cdot (0.7)^{11} \] \[ \approx 0.1934917633 + 1.0087936 \] \[ \approx 0.1934917633 + 0.360586165 = 0.5540779283 \]
Step 4: Round to four decimal places
Thus, the final answer is: \[ P(X \leq 1) \approx 0.5541 \]
So, the probability \( P(1 \text{ or fewer}) \) is approximately \( \boxed{0.5541} \).