Question 1209659
This is a binomial probability problem. Here's how to solve it:

**1. Define the variables:**

*   n = 12 (number of trials/adults selected)
*   p = 0.17 (probability of success/favoring drones)
*   q = 1 - p = 0.83 (probability of failure/not favoring drones)
*   x = number of successes (adults favoring drones)

**2. Binomial probability formula:**

P(x) = (nCx) * p^x * q^(n-x)

where nCx is the binomial coefficient, calculated as n! / (x! * (n-x)!)

**(a) Exactly three:**

P(x = 3) = (12C3) * (0.17)^3 * (0.83)^9
P(x = 3) = (220) * (0.004913) * (0.185665)
P(x = 3) ≈ 0.201

**(b) At least four:**

This means x = 4, 5, 6, ..., 12.  It's easier to calculate the complement and subtract from 1:

P(x ≥ 4) = 1 - P(x < 4)
P(x ≥ 4) = 1 - [P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)]

*   P(x = 0) = (12C0) * (0.17)^0 * (0.83)^12 ≈ 0.122
*   P(x = 1) = (12C1) * (0.17)^1 * (0.83)^11 ≈ 0.298
*   P(x = 2) = (12C2) * (0.17)^2 * (0.83)^10 ≈ 0.312
*   P(x = 3) ≈ 0.201 (calculated above)

P(x ≥ 4) = 1 - (0.122 + 0.298 + 0.312 + 0.201)
P(x ≥ 4) = 1 - 0.933
P(x ≥ 4) ≈ 0.067

**(c) Less than eight:**

This means x = 0, 1, 2, ..., 7.  Again, it's easier to calculate the complement:

P(x < 8) = 1 - P(x ≥ 8)
P(x < 8) = 1 - [P(x = 8) + P(x = 9) + P(x = 10) + P(x = 11) + P(x = 12)]

Calculating each of these probabilities:

*   P(x = 8) = (12C8) * (0.17)^8 * (0.83)^4 ≈ 0.0004
*   P(x = 9) = (12C9) * (0.17)^9 * (0.83)^3 ≈ 0.00005
*   P(x = 10), P(x = 11), and P(x = 12) will be even smaller and essentially zero for practical purposes.

P(x < 8) ≈ 1 - (0.0004 + 0.00005)
P(x < 8) ≈ 0.9995

So, P(x < 8) is very close to 1.  It's highly likely that fewer than 8 adults in the sample will favor the use of drones.