You seem to be asking about the probability of a binomial distribution, specifically \( P(X = 15) \), where \( n = 16 \) and \( p = 0.7 \).

In a binomial distribution, the probability of getting exactly \( k \) successes (in this case, \( X = 15 \)) is calculated using the formula:

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

Where:
- \( \binom{n}{k} \) is the binomial coefficient \( \frac{n!}{k!(n-k)!} \),
- \( p \) is the probability of success on a single trial (here, \( p = 0.7 \)),
- \( n \) is the number of trials (here, \( n = 16 \)),
- \( k \) is the number of successes (here, \( k = 15 \)).

Let’s calculate this:

1. **Binomial Coefficient**: 
   \[
   \binom{16}{15} = \frac{16!}{15!(16-15)!} = 16
   \]

2. **Probability Term**: 
   \[
   p^{15} = 0.7^{15} 
   \]
   and 
   \[
   (1 - p)^{1} = 0.3^1 = 0.3
   \]

Now, multiplying all these values together will give us the result:

\[
P(X = 15) = 16 \cdot (0.7^{15}) \cdot 0.3
\]

Would you like me to compute the final result?

Learn how to calculate the binomial probability P(X=15) when n=16 and p=0.7. This problem involves the binomial distribution formula and includes a step-by-step explanation, suitable for Grades 10-12.

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