Question 557258
70% of Valorites can answer the previous question correctly, suppose you select 3 Valorites at random, what is the probability that at least one of the 3 can answer correctly?
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Binomial Problem with n = 3 and p(correct) = 0.7
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P(1 <= x <= 3) 
= 1 - P(x=0) 
= 1- 3C0(0.7)^0(0.3)^(3) 
= 1 - 0.3^3 
= 0.973
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Cheers,
Stan H.
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