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This is a binomial distribution probability problem because it meets all four requirements:
1. There are a fixed number of trials (10 patients)
2. The are only two outcomes for each trial (patient either develops diabetes, or they don't)
3. The probability of "success" is the same for each trial (30%)
4. The outcome of each trial is independent (one patient developing diabetes does not affect whether the any other patient does or does not)
The formula to use for this probability is:
P(x) = C(n, x) * (p)^x * (1 - p)^(n - x)
where n is the total number of trials (10), x is the number of successful outcomes (5 in this case, since the question asks for the probability that "half" of the 10 develop diabetes), and p is the probability of "success" for each individual trial (30% = 0.30). C(n, x) is the notation that represents the number of combinations of "x" items chosen from a pool of "n" items.
The probability that 5 of the 10 patients develop diabetes is then:
P(5) = C(10, 5) * (0.3)^5 * (1 - 0.3)^(10 - 5)
P(5) = C(10, 5) * (0.3)^5 * (0.7)^5
P(5) = 252 * 0.0024 * 0.1681
P(5) = 0.1029
Please feel free to ask if you have any questions about this solution.