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If 1% of the population is ambidextrous, what is the probability of there being at least one ambidextrous person in a group of 32 randomly selected people?

Answer :

madethisfortwitch1

Answer:

about .275

Step-by-step explanation:

Let's assume that each trail is independent of the last, making this a binomail distribtuion

At least one = 1-exactly 0

exactly 0 = [tex]1-{32\choose0}(1-.01)^{32}*.01^0=1-.99^{32}=.275019664[/tex]

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