Question: User: c. The probability of exactly 9 successes, P(X = 9)
Using the formula with k = 9:
P(X = 9) = \binom{10}{9} (0.56)^9 (0.44)^{10-9}
1. **State the problem:** Calculate the probability of exactly 9 successes in 10 trials where the success probability is $0.56$.
2. **Formula:** The binomial probability formula is
$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$
where $n=10$, $k=9$, and $p=0.56$.
3. **Calculate the binomial coefficient:**
$$\binom{10}{9} = 10$$
4. **Calculate the powers:**
$$0.56^9 \approx 0.0060466$$
$$0.44^{10-9} = 0.44^1 = 0.44$$
5. **Multiply all parts:**
$$P(X=9) = 10 \times 0.0060466 \times 0.44$$
6. **Intermediate multiplication:**
$$10 \times 0.0060466 = 0.060466$$
7. **Final multiplication:**
$$0.060466 \times 0.44 = 0.026605$$
8. **Answer:**
$$P(X=9) \approx 0.0266$$
This means the probability of exactly 9 successes is approximately $0.0266$.