if a player were to take 25 shots in a game what is probability that they would miss at least three if they were to shoot 80%

To solve this problem, we can use the **binomial distribution**. The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success.

### Given Data:
- The probability of making a shot (success) = \( p = 0.80 \)
- The probability of missing a shot (failure) = \( q = 1 - p = 0.20 \)
- Number of shots taken = \( n = 25 \)
- We need the probability of missing **at least three shots**.

### Steps to Solve:

1. **Find the probability of missing fewer than three shots**: This means either missing 0, 1, or 2 shots.
   
2. **Calculate the probabilities**:
   - Probability of missing 0 shots: \( P(X = 0) \)
   - Probability of missing 1 shot: \( P(X = 1) \)
   - Probability of missing 2 shots: \( P(X = 2) \)

   We use the binomial probability formula:
   \[
   P(X = k) = \binom{n}{k} p^{n-k} q^k
   \]
   Where \( k \) is the number of misses.

3. **Sum these probabilities** to find the total probability of missing fewer than three shots.

4. **Subtract this sum from 1** to get the probability of missing at least three shots.

### Calculations:

#### 1. Probability of missing 0 shots:
\[
P(X = 0) = \binom{25}{0} (0.80)^{25-0} (0.20)^0 = (1)(0.80)^{25} \approx 0.0032
\]

#### 2. Probability of missing 1 shot:
\[
P(X = 1) = \binom{25}{1} (0.80)^{24} (0.20)^1 = 25 \times 0.80^{24} \times 0.20 \approx 0.0332
\]

#### 3. Probability of missing 2 shots:
\[
P(X = 2) = \binom{25}{2} (0.80)^{23} (0.20)^2 = 300 \times 0.80^{23} \times 0.04 \approx 0.1076
\]

#### 4. Sum of these probabilities:
\[
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.0032 + 0.0332 + 0.1076 = 0.144
\]

#### 5. Probability of missing at least 3 shots:
\[
P(X \geq 3) = 1 - P(X < 3) = 1 - 0.144 = 0.856
\]

### Conclusion:
The probability that the player would miss at least 3 shots out of 25, given an 80% shooting accuracy, is approximately **0.856** or **85.6%**.

Would you like more details on this calculation, or do you have any other questions?

Here are some related questions:
1. How does the binomial distribution work for different values of \( p \) and \( n \)?
2. How would the probability change if the player took 50 shots instead of 25?
3. Can you explain how the binomial distribution relates to the normal distribution for large \( n \)?
4. What is the probability of making exactly 20 shots out of 25?
5. How does this calculation change if the shooting percentage is not constant?

**Tip**: When working with binomial distributions, a calculator with statistical functions can speed up these probability calculations, especially for large numbers.

Learn how to calculate the probability that a player misses at least three shots out of 25, given an 80% shooting accuracy. This problem uses the binomial distribution to model independent shot attempts with a constant probability of success. Suitable for high school students studying probability and statistics.

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