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Mathematics
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At a local University 54.3% of incoming first-year students have computers. If 3 students are selected at random, find the following probability. A) At least one has a computer B) None have computers
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.543^3=a .457^3=b
that's the wrong answer for A), thanks for B, helps alot. the answer for A is .9046, don't know how i got that? any suggestions
oh only one has to have it. my bad
what does that mean, sorry, how would I get that .9046?
should be 1.0 - the answer for b. only one has to have one, but they can all have one. i ran it as if they all had to have one.
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thanks a ton!
welcome
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