How would you classify the number 121?
1 answer:
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Answer:
The probability that at least one of the children get the disease from their mother is 0.7125.
Step-by-step explanation:
We are given that a human gene carries a certain disease from the mother to the child with a probability rate of 34%.
Suppose a female carrier of the gene has three children. Assume that the infections of the three children are independent of one another.
Let Probability that children get the disease from their mother = P(A) = 0.34
SO, Complement of the event "At least one of the children get the disease from their mother"= P(A') = 1 - P(A)
where A' = event that children do not get the disease from mother.
So, P(A') = 1 - P(A) = 1 - 0.34 = 0.66
Now, probability that at least one of the children get the disease from their mother = 1 - Probability that none of the three children get disease from their mother
= 1 - P(X = 0)
= 1 - (0.66 0.66
0.66)
= 1 - 0.2875 = <u>0.7125</u>