A chance of pulling a character is equal to 0,8%. I make 11 attempts. What are the chances of me getting 3 characters?

To solve this problem, we need to use the binomial probability formula. The probability of getting exactly \( k \) successes (pulling a character) in \( n \) trials (attempts) is given by:

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

where:
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \)
- \( n \) is the number of trials (attempts), which is 11
- \( k \) is the number of successes (pulling a character), which is 3
- \( p \) is the probability of success on a single trial, which is 0.008
- \( 1-p \) is the probability of failure on a single trial

Let's compute it step by step:

1. Calculate the binomial coefficient \( \binom{11}{3} \):
\[
\binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11!}{3! \cdot 8!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165
\]

2. Calculate \( p^3 \):
\[
p^3 = (0.008)^3 = 0.008 \times 0.008 \times 0.008 = 0.000000512
\]

3. Calculate \( (1-p)^{11-3} = (1-0.008)^8 \):
\[
(1-0.008)^8 = 0.992^8 \approx 0.93646
\]

4. Multiply these values together:
\[
P(X = 3) = 165 \times 0.000000512 \times 0.93646 \approx 0.00007944
\]

Therefore, the probability of getting exactly 3 characters in 11 attempts is approximately 0.007944, or about 0.7944%.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:
1. What are the chances of getting at least one character in 11 attempts?
2. What is the expected number of characters pulled in 11 attempts?
3. How would the probability change if the chance of pulling a character increased to 1%?
4. What are the chances of getting no characters in 11 attempts?
5. How does the binomial distribution compare to a Poisson distribution for rare events?

**Tip:** When dealing with small probabilities and many trials, approximations like the Poisson distribution can sometimes simplify calculations.

Learn how to calculate the probability of getting exactly 3 characters in 11 attempts using the binomial probability formula. Step-by-step solution with detailed computations and explanations provided.

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