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(A) This is a binomial distribution problem where $n = 6$ and $p = 10\% = 0.1$.The probability of success (death) is $p = \frac{1}{10}$,and the probab

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$1458 	imes 10^{-5}$

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(A) This is a binomial distribution problem where $n = 6$ and $p = 10\% = 0.1$.The probability of success (death) is $p = \frac{1}{10}$,and the probability of failure (survival) is $q = 1 - p = \frac{9}{10}$.The probability of $x$ successes in $n$ trials is given by the formula $P(X = x) = {}^nC_x p^x q^{n-x}$.We need to find the probability that exactly $3$ patients die,so $x = 3$.$P(X = 3) = {}^6C_3 \left(\frac{1}{10}ight)^3 \left(\frac{9}{10}ight)^3$Calculating the values:${}^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$$P(X = 3) = 20 	imes \left(\frac{1}{1000}ight) 	imes \left(\frac{729}{1000}ight)$$P(X = 3) = 20 	imes \frac{729}{1000000} = \frac{14580}{1000000} = 1458 	imes 10^{-5}$.

The records of a hospital show that $10\%$ of the cases of a certain disease are fatal. If $6$ patients are suffering from the disease,then the probability that only three will die is

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