To solve this problem, we are asked to find the probability $p$ for a binomial distribution $X \sim \text{Bin}(n, p)$, where $n = 5$ and $P(X = 0) = 0.00032$. Let's walk through the steps:

Step 1: Understanding the Binomial Distribution

The probability mass function (PMF) of a binomial distribution is given by the formula:

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

Where:

• $P(X = k)$ is the probability of observing exactly $k$ successes.
• $\binom{n}{k}$ is the binomial coefficient, calculated as $\frac{n!}{k!(n-k)!}$.
• $n$ is the number of trials.
• $p$ is the probability of success in each trial.
• $(1 - p)$ is the probability of failure.

Since we're given $P(X = 0)$, we need to compute this for $k = 0$:

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

This simplifies to:

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

Step 2: Using the Given Probability

We know that $P(X = 0) = 0.00032$, so we substitute this into the equation:

$(1 - p)^5 = 0.00032$

Step 3: Solving for $p$

To solve for $p$, we first take the 5th root of both sides to eliminate the exponent:

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

Next, calculate the 5th root of 0.00032:

$1 - p = 0.2$

Now, solve for $p$:

$p = 1 - 0.2 = 0.8$

Final Answer:

The probability $p$ is $0.8$.

Conclusion:

The probability of success in each trial $p$ for the binomial distribution is 0.8, given that $P(X = 0) = 0.00032$ when $n = 5$.

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Do you want more details on any specific step or have any questions?

Here are 5 related questions for further exploration:

1. How does the binomial distribution change when $p$ increases or decreases?
2. What would $P(X = 1)$ be for $p = 0.8$ and $n = 5$?
3. How do you calculate the mean and variance of a binomial distribution?
4. What is the relationship between the binomial and normal distributions?
5. How would the problem change if $P(X = 5)$ were given instead of $P(X = 0)$?

Tip: In binomial distributions, when $p$ is close to 1, most of the probability mass is concentrated at higher values of $X$, meaning you're more likely to observe a larger number of successes.