\begin{tabular}{l} Question \\ A weighted coin has a 0.55 probability of landing on heads. If you toss the coin 14 times, what is the probability of getting \\ heads exactly 9 times? (Round your answer to 3 decimal places if necessary.) \\ Sorry, that's incorrect. Try again? \\ \hline\end{tabular}

To find the probability of getting exactly 9 heads when flipping a weighted coin 14 times, we can use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where: - \( n = 14 \) (total number of coin flips) - \( k = 9 \) (number of heads we want) - \( p = 0.55 \) (probability of getting heads) - \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \). First, calculate the binomial coefficient \(\binom{14}{9}\): \[ \binom{14}{9} = \frac{14!}{9!5!} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002 \] Now substitute back into the formula to get \( P(X = 9) \): \[ P(X = 9) = 2002 \times (0.55)^9 \times (0.45)^{5} \] Calculating the powers: \[ (0.55)^9 \approx 0.002305 \] \[ (0.45)^5 \approx 0.018578 \] Now plug these values into the probability formula: \[ P(X = 9) \approx 2002 \times 0.002305 \times 0.018578 \] Calculation: \[ P(X = 9) \approx 2002 \times 0.0000427 \approx 0.0856 \] Thus, rounding to 3 decimal places, the probability of getting heads exactly 9 times is approximately: \[ \mathbf{0.086} \]