Answer: Approximately 0.026673
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Step-by-step explanation:
n = 20 is the sample size
p = 0.035 is the probability of selecting someone who is from multiple births (twins, triplets, etc)
Either someone is a twin/triplet/etc or they aren't. This two option nature is a strong implication we have a binomial probability.
The binomial probability formula is
B(x) = (n C x)*(p)^x*(1-p)^(n-x)
where x is the number of successes and the n C x refers to the nCr combination formula.
We want to know the value of B(3), ie the value of B(x) when we have exactly x = 3 children from multiple births.
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B(x) = (n C x)*(p)^x*(1-p)^(n-x)
B(3) = (20 C 3)*(0.035)^3*(1-0.035)^(20-3)
B(3) = (1140)*(0.035)^3*(0.965)^(17)
B(3) = 0.026673
There's roughly a 2.6673% chance of the 20 person sample having exactly 3 children who are from multiple births.
Side note: You can compute 20C3 = 1140 using the conventional nCr formula, or you can use Pascal's Triangle as an alternative route.