This problem deals with a binomial probability scenario.

Given:

• Probability (p) that a person uses a smartphone in meetings or classes: $p = 0.47$
• Number of trials (n): $n = 10$ (since 10 adults are selected)
• We need to find the probability that at least 5 of the selected adults use their smartphones in meetings or classes.

Solution Strategy:

To calculate the probability of "at least 5" using their smartphones, we'll sum the probabilities for 5, 6, 7, 8, 9, and 10 using the binomial probability formula:

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

• $n$ is the total number of trials (10 in this case),
• $k$ is the number of successes (number of people using smartphones),
• $p$ is the probability of success (0.47),
• $1 - p = 0.53$ is the probability of failure.

Instead of calculating each probability manually, a more efficient way is to calculate the complement of the event. The complement of "at least 5" is "fewer than 5." So:

$P(X \geq 5) = 1 - P(X < 5)$

Thus, we calculate $P(X < 5)$, which is the sum of the probabilities for 0, 1, 2, 3, and 4 people using their smartphones.

Let me compute this for you:The probability that at least 5 of the 10 selected adults use their smartphones in meetings or classes is 0.5474 (rounded to four decimal places).

Would you like further details or have any questions? Here are five related questions you might explore:

1. What is the probability that exactly 5 of the adults use their smartphones in meetings or classes?
2. How does the probability change if only 8 adults are selected instead of 10?
3. What if the percentage of adults using smartphones changes to 60%? How does this affect the outcome?
4. Can this problem be modeled using a normal approximation?
5. What is the expected number of adults using smartphones if 10 adults are selected?

Tip: When dealing with binomial probabilities, calculating the complement (like we did here) often simplifies the process for "at least" or "more than" cases.