Given :\(33\frac{1}{3}\%\) of the workers have a chance of suffering from the diseaseConcept:The formula for binomial distribution:\(P\left( x \right) = \;{n_{{C_x}}}{p^x}{q^{n - x}}\)where,p = probability of success of a trialq = (1- p) = probability of failure of a trailn = total no of trialx = no of a successful trial for the given condition.\({n_{{C_x}}} = \frac{{n!}}{{\left( {n - x} \right)!x!}}\) = number of combinations of n taken x at a timeP(x) = probability of success of x trialCalculation:From the previous question,The probability that no one out of 6 workers suffers from a diseaseP(A) = \(\displaystyle \frac{64}{729}\)Let, P(B) is the probability that at least one out of 6 workers suffer from a disease.We know that, P(A) + P(B) = 1? P(B) = 1 - p(A) = 1 - \(\displaystyle \frac{64}{729}\)? p(B) = \(\displaystyle \frac{665}{729}\)Alternate Methodp = \(\frac{1}{3}\) & q = 1 - p = 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)Probability, where out of 6, exactly 3 suffer from the disease isP (X ? 1)? P (X ? 1) = P(x = 1) + (x = 2) + (x = 3) + (x = 4) + (x = 5) + (x = 6)? P (X ? 1) = \(\displaystyle ^6C_1\ (\frac{1}{3})^1\ (\frac{2}{3})^{6-1}+\ ^6C_2\ (\frac{1}{3})^2\ (\frac{2}{3})^{6-2}+\ ^6 C_3\ (\frac{1}{3})^3\ (\frac{2}{3})^{6-3}+ \ ^6C_4\ (\frac{1}{3})^4\ (\frac{2}{3})^{6-4}+\ ^6C_5\ (\frac{1}{3})^5\ (\frac{2}{3})^{6-5}+\ ^6C_6\ (\frac{1}{3})^6\ (\frac{2}{3})^{6-6}\) ? P (X ? 1) = \(\displaystyle ^6C_1\ \frac{{2}^5}{{3}^6}+\ ^6C_2\ \frac{{2}^4}{{3}^6}+\ ^6 C_3\ \frac{{2}^3}{{3}^6}+ \ ^6C_4\ \frac{{2}^2}{{3}^6}+\ ^6C_5\ \frac{{2}^1}{{3}^6}+\ ^6C_6\ \frac{{2}^0}{{3}^6}\) ? P (X ? 1) = \(\displaystyle \frac{6!}{5!.1!} \frac{{2}^5}{{3}^6}+ \frac{6!}{4!.2!} \frac{{2}^4}{{3}^6}+ \frac{6!}{3!.3!} \frac{{2}^3}{{3}^6}+ \frac{6!}{2!.4!} \frac{{2}^2}{{3}^6}+ \frac{6!}{1!.5!} \frac{{2}^1}{{3}^6}+ \frac{6!}{6!} \frac{{2}^0}{{3}^6}\)? P (X ? 1) = \(\displaystyle \frac{665}{729}\)? The probability that at least one out of 6 workers suffer from a disease is \(\displaystyle \frac{665}{729}\)