Solution to Question d:

The problem involves a binomial probability distribution since it deals with a fixed number of trials, two possible outcomes (effective or not effective), and a constant probability of success.

Step 1: Identify the variables

• $n = 10$: Number of patients
• $p = 0.80$: Probability that the medication is effective for a single patient
• $x = 7$: Number of patients for whom the medication is effective
• $q = 1 - p = 0.20$: Probability that the medication is not effective

The formula for the binomial probability is:

$P(X = x) = \binom{n}{x} p^x q^{n-x}$

Step 2: Compute the binomial coefficient

The binomial coefficient $\binom{n}{x}$ is given by:

$\binom{n}{x} = \frac{n!}{x! (n-x)!}$

$\binom{10}{7} = \frac{10!}{7! (10-7)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

Step 3: Plug values into the formula

Now substitute $n = 10$, $x = 7$, $p = 0.80$, and $q = 0.20$ into the formula:

$P(X = 7) = \binom{10}{7} (0.80)^7 (0.20)^3$

1. Compute $(0.80)^7$: $(0.80)^7 = 0.2097152$

2. Compute $(0.20)^3$: $(0.20)^3 = 0.008$

3. Multiply the probabilities: $0.2097152 \times 0.008 = 0.0016777216$

4. Multiply by the binomial coefficient: $P(X = 7) = 120 \times 0.0016777216 = 0.20133$

Final Answer:

The probability that the medication is effective in exactly 7 out of 10 patients is approximately:

$P(X = 7) \approx 0.201$

Would you like further clarification or help with the other parts? 😊

Related Questions:

1. How is the binomial distribution applied to real-world problems?
2. What would be the probability if the medication were effective for 8 patients instead of 7?
3. How can cumulative probabilities for $X \geq 7$ be calculated?
4. Can this problem be solved using normal approximation? If so, how?
5. What are the properties of a binomial random variable?

Tip: Always double-check your calculations to avoid rounding errors, especially with probabilities!