The occurrence of a disease in an industry is such that the workers have 20% chance of suffering from it. What is the probability that out of 6 workers chosen at random, 4 or more will suffer from the disease?

Explanation

This is a binomial distribution with n=6, p=1/5, and q=4/5. The probability of 4 or more suffering is P(X \geq 4) = P(X=4) + P(X=5) + P(X=6). This is \binom{6}{4}(\frac{1}{5})^4(\frac{4}{5})^2 + \binom{6}{5}(\frac{1}{5})^5(\frac{4}{5})^1 + \binom{6}{6}(\frac{1}{5})^6(\frac{4}{5})^0. Evaluating gives \frac{15 \times 16}{15625} + \frac{6 \times 4}{15625} + \frac{1}{15625} = \frac{240 + 24 + 1}{15625} = \frac{265}{15625} = \frac{53}{3125}.

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