If the probability that an individual stuffers a bad reaction from a certain injection is 0.001, determine the a) Exactly 3  b)more than 2 c)none d)more than 1 individual suffer a

This problem is best solved using the **binomial probability formula**, which is typically used when there are only two possible outcomes (success or failure) for each trial. In this case:

- **Probability of a bad reaction (success)**: \( p = 0.001 \)
- **Probability of no bad reaction (failure)**: \( q = 1 - p = 1 - 0.001 = 0.999 \)
- Let's assume that \( n \) individuals are injected.

### Binomial Probability Formula:

The probability of exactly \( k \) successes (bad reactions) in \( n \) trials is given by:

\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}
\]

Where:
- \( X \) is the number of individuals who suffer a bad reaction.
- \( \binom{n}{k} \) is the binomial coefficient.
- \( p^k \) is the probability of \( k \) bad reactions.
- \( (1 - p)^{n-k} \) is the probability that the rest don't suffer a bad reaction.

Let’s break down each part of the question:

### a) Probability that **exactly 3** individuals suffer a bad reaction:

This can be calculated using the binomial probability formula for \( k = 3 \). The probability would be:

\[
P(X = 3) = \binom{n}{3} \cdot p^3 \cdot (1 - p)^{n - 3}
\]

### b) Probability that **more than 2** individuals suffer a bad reaction:

This can be expressed as:

\[
P(X > 2) = 1 - P(X \leq 2)
\]

So, you need to subtract the cumulative probability for 0, 1, and 2 individuals suffering from the total probability (which is 1).

### c) Probability that **none** of the individuals suffer a bad reaction:

This is simply \( P(X = 0) \), which can be calculated as:

\[
P(X = 0) = \binom{n}{0} \cdot p^0 \cdot (1 - p)^n = (1 - p)^n
\]

### d) Probability that **more than 1** individual suffers a bad reaction:

This is similar to part b, but here we calculate:

\[
P(X > 1) = 1 - P(X \leq 1)
\]

Where \( P(X \leq 1) \) is the cumulative probability for 0 and 1 bad reactions.

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To proceed with these calculations, I would need the **number of individuals** (denoted by \( n \)) that are being injected. Could you provide that value so I can calculate the probabilities more precisely?

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Do you want details or have any questions?

Here are 5 questions related to this problem:

1. How do we calculate the binomial coefficient \( \binom{n}{k} \)?
2. Why is the binomial probability distribution suitable for this problem?
3. What is the significance of the complement rule \( P(X > 2) = 1 - P(X \leq 2) \)?
4. How does the probability change if the value of \( p \) increases or decreases?
5. What are the assumptions made in a binomial probability problem?

**Tip**: Always double-check whether your problem requires "exact" probabilities or cumulative ones (like "more than" or "less than"). This ensures you're solving the right question!

If the probability that an individual suffers a bad reaction from a certain injection is 0.001, determine the a) Exactly 3 b) more than 2 c) none d) more than 1 individual suffer a

Explore how to calculate the binomial probability for exactly 3, more than 2, none, or more than 1 individual suffering a bad reaction from an injection with a probability of 0.001. This problem involves key probability concepts and binomial distribution, suitable for advanced high school or college-level students.

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