Question: A wildlife researcher is analyzing the genetic diversity of a population of 10 endangered gorillas. If 4 of them are tagged with GPS collars, and the researcher selects 3 gorillas at random for detailed monitoring, what is the probability that at least one of the selected gorillas has a GPS collar?

["Understanding the Probability of Selecting GPS-Collar-Tagged Gorillas in Conservation Research", "In wildlife conservation, understanding genetic diversity within endangered populations is vital for developing effective protection and breeding strategies. A recent study by a wildlife researcher examining a population of 10 endangered gorillas highlights an important statistical question: If 4 out of 10 gorillas are tagged with GPS collars and 3 are randomly selected for detailed monitoring, what is the probability that at least one of the selected gorillas is GPS-tagged?", "This scenario invites an analysis of probability using combinatorial methods, crucial for accurate ecological inference.", "### Population and Selection Setup", "- Total gorillas: 10\n- GPS-collared gorillas: 4\n- Non-collared gorillas: 10 – 4 = 6\n- Gorillas selected for monitoring: 3 (chosen at random without replacement)", "We are asked to find the probability that at least one of the 3 selected gorillas is GPS-tagged.", "### Probability Strategy: Use of Complementary Probability", "Calculating “at least one” directly involves multiple cases (exactly 1, 2, or 3 GPS-collared gorillas). A simpler alternative is using complementary probability:", "[\nP(\ ext{at least one tagged}) = 1 - P(\ ext{none tagged})\n]", "### Calculating the Probability That None Are Tagged", "To select 3 gorillas with none collared:", "- Choose all 3 from the 6 non-tagged gorillas\n- Total ways to choose 3 gorillas from 10:\n[\n\binom{10}{3} = \frac{10 \ imes 9 \ imes 8}{3 \ imes 2 \ imes 1} = 120\n]", "- Ways to choose 3 non-collared gorillas from 6:\n[\n\binom{6}{3} = \frac{6 \ imes 5 \ imes 4}{3 \ imes 2 \ imes 1} = 20\n]", "So, the probability that none of the selected gorillas are tagged is:\n[\nP(\ ext{none tagged}) = \frac{\binom{6}{3}}{\binom{10}{3}} = \frac{20}{120} = \frac{1}{6}\n]", "### Final Probability Calculation", "Now compute the desired probability:\n[\nP(\ ext{at least one tagged}) = 1 - \frac{1}{6} = \frac{5}{6}\n]", "### Interpretation and Research Implications", "The probability that at least one of the 3 randomly selected gorillas carries a GPS collar is (\frac{5}{6}), or approximately 83.3%. This result indicates a high likelihood that GPS tracking data is included in any random selection, making the monitoring effort representative and reducing the risk of bias in genetic diversity assessments.", "For wildlife researchers, understanding such probabilities ensures robust sampling strategies, critical for accurate analysis of endangered species’ genetic health and guiding conservation decisions.", "### Conclusion", "In conservation genetics, strategic sampling is essential. By applying combinatorial probability, researchers can estimate the chances of capturing tagged individuals, enhancing the reliability of fieldwork and supporting targeted conservation actions for critically endangered species like the gorillas studied.", "---", "Keywords: gorilla conservation, genetic diversity, GPS collar, probability calculation, wildlife research, sampling probability, endangered species monitoring, complementary probability, wildlife statistics, conservation genetics."]