p = 83% = 0.83
n = 21
Binomial probability:
"P(X=x)=C(n,x)\\cdot p^x\\cdot (1-p)^{n-x}"
"P(X\\le1)=P(X=0)+P(X=1)"
"P(X\\le1)=\\frac{21!}{0!21!}\\cdot 0.83^0\\cdot 0.17^{21}+\\frac{21!}{1!20!}\\cdot 0.83^1\\cdot 0.17^{20}=7.15304728\\cdot10^{\u221215}"
The probability that no more than 1 of the 21 adults require eyesight correction is close to zero. Yes, 1 is a significantly low number of adults requiring eyesight correction.
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