Based on a Comcast survey, there is a 0.8 probability that a randomly selected adult will watch prime-time TV live, instead of online, on DVR, etc. Assume that seven adults are randomly selected. Find the probability that fewer than three of the selected adults watch prime-time live.
A. 0.00467
B. 0.000358
C. 0.00430
D. 0.0512

Step 1 ：This problem is about the binomial distribution. We are given the probability of success (\(p = 0.8\)), the number of trials (\(n = 7\)), and we need to find the probability of having fewer than three successes (\(k < 3\)). The formula for the binomial distribution is: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where \(C(n, k)\) is the combination of n items taken k at a time.

Step 2 ：However, since we need to find the probability of having fewer than three successes, we need to sum up the probabilities for \(k = 0\), \(1\), and \(2\).

Step 3 ：The calculation gives us a probability of approximately 0.00467.

Step 4 ：Final Answer: The probability that fewer than three of the selected adults watch prime-time live is approximately \(\boxed{0.00467}\).