If there is a 1/30,000 chance of something happening and a 1/10,000 chance of something happening. Then what are the chances of them happening within 100 encounters of eachother

To calculate the probability of two independent events happening within 100 encounters, where each event has a separate chance of occurring in each encounter, we need to understand that we’re dealing with two events occurring in the same range (within 100 trials). Let's denote these events as follows:

- Event A: has a \( \frac{1}{30,000} \) chance of happening per encounter.
- Event B: has a \( \frac{1}{10,000} \) chance of happening per encounter.

### Step 1: Probability of Each Event Happening Within 100 Encounters
For each event, we calculate the probability that it happens at least once within 100 encounters.

#### Probability for Event A
The probability of Event A **not** happening in a single encounter is:
\[
1 - \frac{1}{30,000} = \frac{29,999}{30,000}
\]

The probability of Event A not happening in 100 encounters is:
\[
\left( \frac{29,999}{30,000} \right)^{100}
\]

So, the probability of Event A happening at least once within 100 encounters is:
\[
P(A \text{ in 100 encounters}) = 1 - \left( \frac{29,999}{30,000} \right)^{100}
\]

#### Probability for Event B
Similarly, the probability of Event B **not** happening in a single encounter is:
\[
1 - \frac{1}{10,000} = \frac{9,999}{10,000}
\]

The probability of Event B not happening in 100 encounters is:
\[
\left( \frac{9,999}{10,000} \right)^{100}
\]

So, the probability of Event B happening at least once within 100 encounters is:
\[
P(B \text{ in 100 encounters}) = 1 - \left( \frac{9,999}{10,000} \right)^{100}
\]

### Step 2: Probability of Both Events Happening Within 100 Encounters
Since we want both events to happen within the same 100 encounters, we assume that these events are independent. Therefore, we can multiply their individual probabilities:

\[
P(\text{both A and B in 100 encounters}) = P(A \text{ in 100 encounters}) \times P(B \text{ in 100 encounters})
\]

Substituting in the probabilities from above, we get:

\[
P(\text{both A and B in 100 encounters}) = \left( 1 - \left( \frac{29,999}{30,000} \right)^{100} \right) \times \left( 1 - \left( \frac{9,999}{10,000} \right)^{100} \right)
\]

### Step 3: Calculating the Result
To provide a more precise answer, let's calculate the values of these probabilities. Here are the approximate values:

1. \( \left( \frac{29,999}{30,000} \right)^{100} \approx 0.9967 \), so \( P(A \text{ in 100 encounters}) \approx 1 - 0.9967 = 0.0033 \).
2. \( \left( \frac{9,999}{10,000} \right)^{100} \approx 0.9900 \), so \( P(B \text{ in 100 encounters}) \approx 1 - 0.9900 = 0.0100 \).

Then,

\[
P(\text{both A and B in 100 encounters}) \approx 0.0033 \times 0.0100 = 0.000033
\]

### Conclusion
The probability of both events occurring within the same 100 encounters is approximately **0.000033** or **0.0033%**.

Would you like further details or have any questions?

---

Here are some follow-up questions for more exploration:
1. How would the probability change if there were 1,000 encounters instead of 100?
2. What is the probability of at least one of the two events occurring within 100 encounters?
3. How does increasing the probability of one event impact the combined probability?
4. What if the two events were not independent? How would that affect the calculation?
5. How would this approach differ if there were three independent events?

**Tip:** In probability, for events with a small chance, the chance of at least one occurrence can be approximated by \(1 - e^{-np}\) where \(p\) is the probability of a single event, and \(n\) is the number of trials.

If there is a 1/30,000 chance of something happening and a 1/10,000 chance of something happening. Then what are the chances of them happening within 100 encounters of each other?

Discover how to calculate the probability of two independent events occurring within 100 encounters. This problem involves a 1/30,000 chance for Event A and a 1/10,000 chance for Event B, resulting in an approximate combined probability of 0.000033 or 0.0033%. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation