Let's define the problem mathematically:

• Suppose each package independently arrives on time with probability $p$.
• The probability of a package not arriving on time is $q = 1 - p$.
• We have 10 packages, so this follows a binomial distribution:
  $X \sim \text{Bin}(10, q)$ where $X$ represents the number of packages that do not arrive on time.

We need to find the probability that at most one package is delayed:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

Using the binomial probability formula:

$P(X = k) = \binom{n}{k} q^k (1-q)^{n-k}$

For $X = 0$ (no package is delayed):

$P(X = 0) = \binom{10}{0} q^0 (1-q)^{10} = (1-q)^{10}$

For $X = 1$ (one package is delayed):

$P(X = 1) = \binom{10}{1} q^1 (1-q)^9 = 10q(1-q)^9$

Thus, the total probability is:

$P(X \leq 1) = (1-q)^{10} + 10q(1-q)^9$

Next Steps:

• If you provide the probability $p$ (or $q$), I can calculate the exact value.
• Let me know if you want an explanation on binomial distributions.

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Here are five related questions:

1. What is the probability that exactly two packages will not arrive on time?
2. How does the probability change if the number of packages increases to 20?
3. What is the expected number of packages that do not arrive on time?
4. How do we approximate this probability using a normal distribution?
5. How can we use the Poisson distribution to approximate this scenario?

Tip:
In large binomial problems, the normal approximation can simplify probability calculations.