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.