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:p = \(\frac{1}{3}\) & q = 1 - p = 1 - \(\frac{1}{3}\) = \(\frac{2}{3}\)Probability, where out of 6, exactly 3 suffer from the disease isP (X = 3)? P(x = 3) = \(\displaystyle ^6C_3\ (\frac{1}{3})^3\ (\frac{2}{3})^{6-3}\)? P(x = 3) = \(\displaystyle ^6C_3\ (\frac{1}{3})^3\ (\frac{2}{3})^3\)? P(x = 3) = \(\displaystyle \frac{6!}{3!.3!} \frac{{2}^3}{{3}^6}\)? P(x = 3) = \(\displaystyle \frac{160}{729}\)? The probability that exactly 3 out of 6 workers suffer from a disease is \(\displaystyle \frac{160}{729}\)