Question

A survey showed that ​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If adults are randomly​ selected, find the probability that of them need correction for their eyesight.

Answer 1

$p = 83\% \\= 0.83\\ n = 21$

Binomial probability:

$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{n-x}$

$P(X\le1)=P(X=0)+P(X=1)$

$P(X\le1)=\frac{21!}{0!21!}\cdot 0.83^0\cdot 0.17^{21}+\frac{21!}{1!20!}\cdot 0.83^1\cdot 0.17^{20}=7.15304728\cdot10^{−15}$

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.