This problem involves a binomial distribution, where the probability of $k$ successes in $n$ trials is given by the binomial formula:

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

Given:

• $n = 5$ (5 independent trials)
• Probabilities for 1 and 2 successes are:
  • $P(X = 1) = 0.4096$
  • $P(X = 2) = 0.2048$

We need to find the parameter $p$, which is the probability of success in a single trial.

Step 1: Formula for 1 success $P(X = 1)$

Using the binomial formula for $X = 1$: $P(X = 1) = \binom{5}{1} p (1 - p)^4$ $0.4096 = 5p (1 - p)^4$

Step 2: Formula for 2 successes $P(X = 2)$

Similarly, for $X = 2$: $P(X = 2) = \binom{5}{2} p^2 (1 - p)^3$ $0.2048 = 10 p^2 (1 - p)^3$

Step 3: Solve for $p$

Now we have two equations: $5p (1 - p)^4 = 0.4096$ $10 p^2 (1 - p)^3 = 0.2048$

By dividing these two equations, we can eliminate some terms and solve for $p$:

$\frac{5p (1 - p)^4}{10 p^2 (1 - p)^3} = \frac{0.4096}{0.2048}$ $\frac{(1 - p)}{2p} = 2$ Multiplying both sides by $2p$, we get: $1 - p = 4p$ $1 = 5p$ $p = 0.2$

Thus, the parameter $p$ is $\boxed{0.20}$, corresponding to option $\text{c}$.

Would you like more details or have any other questions?

Here are 5 related questions for further understanding:

1. How is the binomial distribution derived?
2. What is the role of the binomial coefficient in this context?
3. How would the probabilities change if the number of trials $n$ increased?
4. Can the binomial formula be used for continuous data?
5. How would you calculate the expected value of this binomial distribution?

Tip: In binomial problems, always carefully match the formula with the given probability for a certain number of successes!