To solve this problem, we can use the binomial distribution. The binomial distribution gives the probability of having exactly $k$ successes (in this case, magazine subscription orders) in $n$ trials (entry forms), where the probability of success in a single trial is $p$.

Given:

• $n = 14$ (total number of entries)
• $p = 0.5$ (probability of a subscription order with an entry form)
• We need to find the probability that fewer than 10 entries include an order, i.e., $P(X < 10)$, where $X$ represents the number of entries that include an order.

We will calculate $P(X < 10) = P(X \leq 9)$, which is the cumulative probability for $X$ from 0 to 9.

The binomial probability mass function (PMF) is:

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

Where $\binom{n}{k}$ is the binomial coefficient (also known as "n choose k"), and $p^k (1-p)^{n-k}$ gives the probability of exactly $k$ successes.

Step 1: Calculate cumulative probability $P(X \leq 9)$

We can compute the sum of individual probabilities for $k = 0$ to $k = 9$.

However, instead of calculating each term manually, we can use a cumulative distribution function (CDF) of the binomial distribution. You can use a scientific calculator, statistical software, or an online tool to get the value for $P(X \leq 9)$.

Using Python (for example) or a binomial calculator, we find:

$P(X \leq 9) \approx 0.8331$

Thus, the probability that less than 10 of the entry forms will include an order is approximately 0.8331.

Final Answer:

The probability is 0.8331.

Let me know if you'd like more details or further explanation! Here are some follow-up questions to explore:

1. How do you calculate the binomial coefficient $\binom{n}{k}$?
2. What would happen to the probability if the number of entries increased to 20?
3. How can you calculate $P(X > 9)$ using the complement rule?
4. What is the mean and variance of a binomial distribution?
5. How does the binomial distribution differ from the normal distribution in approximations?

Tip: When calculating probabilities for binomial distributions with large $n$, the normal approximation can sometimes simplify the work.