To determine the probability that more than 7 out of 10 people will buy a new iPhone after viewing an advertisement, given that historically 30% of people who see the ad make a purchase, we can model this situation using a binomial distribution.

Problem Setup:

• n = 10 (the number of people who view the ad),
• p = 0.30 (the probability that a single person buys the iPhone after viewing the ad),
• We need to find the probability that more than 7 people (i.e., 8, 9, or 10 people) buy the iPhone.

Binomial Probability Formula:

For a binomial distribution, the probability of getting exactly $k$ successes (people buying the iPhone) is given by:

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

Where:

• $\binom{n}{k}$ is the binomial coefficient, which is calculated as $\frac{n!}{k!(n-k)!}$,
• $p$ is the probability of success (in this case, 0.30),
• $n$ is the number of trials (10 people),
• $k$ is the number of successes we are interested in.

Steps to Solve:

We need to find $P(X > 7)$, which means:

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)$

Let’s calculate each term individually. I'll do the computation to sum up these probabilities.The probability that more than 7 out of 10 people will buy a new iPhone after viewing the advertisement is approximately 0.00159, or about 0.16%.

Would you like further details or clarifications?

Here are 5 related questions to expand on this:

1. How does the probability change if 15 people view the advertisement instead of 10?
2. What is the probability that exactly 5 people will buy the iPhone?
3. Can you explain the concept of a binomial distribution in more detail?
4. How would the probability change if the historical data showed 50% of viewers bought the iPhone?
5. How does the binomial distribution relate to the normal distribution?

Tip: In a binomial distribution, as the number of trials increases, the distribution starts to resemble a normal distribution, especially when $np$ and $n(1-p)$ are both large.