The incidence of occupational disease in an i | vedclass

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(C) This is a binomial distribution problem where $n = 5$ and $p = 0.1$ (or $\frac{1}{10}$),so $q = 1 - p = 0.9$ (or $\frac{9}{10}$).We need to find t

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$0.00856$

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(C) This is a binomial distribution problem where $n = 5$ and $p = 0.1$ (or $\frac{1}{10}$),so $q = 1 - p = 0.9$ (or $\frac{9}{10}$).We need to find the probability $P(X \ge 3) = P(X=3) + P(X=4) + P(X=5)$.Using the formula $P(X=k) = {}^nC_k \cdot p^k \cdot q^{n-k}$:$P(X=3) = {}^5C_3 \cdot (\frac{1}{10})^3 \cdot (\frac{9}{10})^2 = 10 \cdot \frac{1}{1000} \cdot \frac{81}{100} = \frac{810}{100000} = 0.00810$$P(X=4) = {}^5C_4 \cdot (\frac{1}{10})^4 \cdot (\frac{9}{10})^1 = 5 \cdot \frac{1}{10000} \cdot \frac{9}{10} = \frac{45}{100000} = 0.00045$$P(X=5) = {}^5C_5 \cdot (\frac{1}{10})^5 \cdot (\frac{9}{10})^0 = 1 \cdot \frac{1}{100000} \cdot 1 = \frac{1}{100000} = 0.00001$Summing these probabilities: $0.00810 + 0.00045 + 0.00001 = 0.00856$.

The incidence of occupational disease in an industry is such that the workmen have a $10 \%$ chance of suffering from it. The probability that out of $5$ workmen,$3$ or more will contract the disease is

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