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We'd like to know the likelihood that any 5 people chosen from 20 will be left-handed if the probability of being left-handed is assumed to be 10%. Well, 5 people can be chosen from 20 in (20 C 5) = 15504 ways. So, we have (15504)*(.1^5)*(.9^15) = 3.19%. This is a Binomial probability problem, where we are interested in 'k' successes (k = 5, success --> people being left-handed) in 'n' trials (n = 20).

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Q: If 10 percent of the population is left handed what is the probability that 5 people in 20 will be left handed?
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