Given that $X \sim \text{Bin}(n, p)$, where $X$ is a binomially distributed random variable with parameters $n = 5$ and $p$ (the probability of success in a single trial), and that $P(X = 0) = 0.00032$, we need to find $p$.

Step-by-Step Solution:

1. Binomial Distribution Formula: The probability mass function (PMF) of a binomial distribution is given by:

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

   Here, $n = 5$, and we are given $P(X = 0) = 0.00032$, so we substitute $k = 0$ into the formula:

   $P(X = 0) = \binom{5}{0} p^0 (1 - p)^5$

2. Simplifying the Formula: Recall that $\binom{5}{0} = 1$ and $p^0 = 1$, so the equation simplifies to:

   $P(X = 0) = (1 - p)^5$

   We are given that $P(X = 0) = 0.00032$, so:

   $(1 - p)^5 = 0.00032$

3. Solving for $p$: To solve for $p$, we first take the fifth root of both sides:

   $1 - p = \sqrt[5]{0.00032}$

   Now, calculate $\sqrt[5]{0.00032}$.

   $1 - p = 0.2$

4. Final Calculation: Solving for $p$:

   $p = 1 - 0.2 = 0.8$

Thus, the probability $p$ is 0.8.

Explanation:

• We used the binomial distribution formula, focusing on the case where $X = 0$.
• After simplifying the binomial coefficient and powers, we reduced the problem to solving the equation $(1 - p)^5 = 0.00032$.
• Taking the fifth root allowed us to isolate $p$, leading to the final solution $p = 0.8$.

Would you like more details or have any questions?

Related Questions:

1. How would the answer change if $P(X = 1)$ were provided instead of $P(X = 0)$?
2. What is the expected value $E(X)$ for this binomial distribution?
3. What is the variance of the binomial distribution $X \sim \text{Bin}(5, 0.8)$?
4. How do you compute the cumulative distribution function (CDF) for a binomial distribution?
5. What happens to the distribution as $n$ increases but $p$ remains constant?

Tip:

When solving binomial distribution problems, always ensure that you simplify factorial and binomial coefficient terms early to avoid unnecessary complexity.