Question: A science fiction writer is designing a universe where each planet has a 1 in 5 chance of hosting intelligent life, independently. If a galactic council selects 4 randomly chosen planets, what is the probability that exactly two of them host intelligent life, given that at least one does?
["How Likely Is It That Exactly Two of Four Planets Host Intelligent Life—With At Least One? \nIn sci-fi speculations and universe-building, a compelling question emerges: What’s the chance that, in a randomly selected cluster of four planets, exactly two teem with intelligent life—each with a 20% independent probability? This calculation, rooted in probability theory, mirrors real-world thinking used in risk modeling, resource forecasting, and even creative probability puzzles. It’s a pattern widely discussed in science communication and speculative thought experiments, making it highly relevant for curious US readers exploring futurism or data-driven storytelling.", "### Why This Question Is Gaining Attention \nWith rising interest in alien life, space colonization, and unpredictable cosmic ecosystems, audiences are increasingly drawn to grasp the statistical fabric of hypothetical universes. The setup—each planet independently hosting life with a 1 in 5 chance—creates a familiar yet rich framework for understanding independent events and conditional probabilities. Platforms like Discover thrive on such questions that blend curiosity and clarity, offering meaningful engagement without speculative overreach.", "### Breaking Down the Probability: Step by Step", "Understanding the Base Model \nEach world independently has a 0.2 (20%) chance of supporting intelligent life. When selecting 4 planets, the full set of outcomes follows a binomial distribution, where success = hosting intelligent life. We want exactly 2 successes (intelligent life) with the added condition: at least one life exists. This shifts the focus from “exactly 2” to “exactly 2 and at least one lives,” a subtle but crucial distinction.", "Calculating “Exactly Two” \nUsing the binomial formula: \n\[ P(X = 2) = \binom{4}{2} (0.2)^2 (0.8)^2 = 6 \ imes 0.04 \ imes 0.64 = 0.1536 \]", "Adjusting for “At Least One” \nThe condition “at least one” rules out total lifelessness (0 planets alive). That probability is: \n\[ P(X = 0) = (0.8)^4 = 0.4096 \] \nSo, the probability of at least one life is: \n\[ 1 -"]
