SOLUTION: A human gene carries a certain disease from the mother to the child with a probability rate of 30%. That is, there is a 30% chance that the child becomes infected with the disease.
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Question 1134902: A human gene carries a certain disease from the mother to the child with a probability rate of 30%. That is, there is a 30% chance that the child becomes infected with the disease. Suppose a female carrier of the gene has five children. Assume that the infections of the five children are independent of one another. Find the probability that at least one of the children get the disease from their mother. Round to the nearest thousandth. Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! Use the binomial probability formula
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Probability (P) (k success in n trials) = nCk * p^k * (1-p)^(n-k), where p is the probability, nCk = n!/(k! * (n-k)!)
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P(at least 1 of the children get the disease) = 1 - P(no children get the disease)
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P(no children get the disease) = 5C0 * (0.30)^0 * (1-0.30)^(5-0) = 0.16807
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P(at least 1 of the children get the disease) = 1 - 0.16807 = 0.83193 is approximately 0.832
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