Answer:
(a) P=0.694
(b) It is independent, beacuse the probability of having the disease for the children depends only on her mother condition (if she has the disease or not), not the condition of his brothers and sisters.
(c) P=0.25
Step-by-step explanation:
(a) If the mother has 0.33 probabilities of having the disease, the probability of the children having the disease is equal to the product of the probability of the mother having the disease (0.33) and the probability of inherit it (0.50).
So the probability of one child of having the disease is 0.33*0.5=0.167. The probability of not having the disease is then (1-0.167)=0.833
The probability of both children to not have the disease is 0.833^2=0.694.
(b) It is independent, beacuse the probability of having the disease for the children depends only on her mother condition (if she has the disease or not), not the ones of his brothers and sisters.
(c) If the mother has the disease, the child have a probability of 0.5 of having the disease.
The probability, given that the mother has the disease, of both child not having the disease is 0.5^2=0.25.
Answer:
C, this diagram will show 2 1/3- 2/3 which is 7/3-2/3 which is 5/3 which is 1 2/3
Step-by-step explanation:
Answer:
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Step-by-step explanation:
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