About 8.3% of the adult population in the US has diabetes. It is known that diab

Question

About 8.3% of the adult population in the US has diabetes. It is known that diabetes may be inherited. Suppose we know that if one of the parents (regardless of gender) has diabetes, the chance that the child develops diabetes up to the age of 40 is 20%. If both parents haw diabetes the chance increases to 40%. If neither parent has diabetes, the chance of the child developing diabetes is 4%. (a) What is the probability that neither one of my parents has din betas? What is the probability that only my dad has diabetes. What is the probability that both have diabetes? (b) Suppose I am an only child and I just discovered that I have the disease. I have no idea whether my dad or mom have diabetes. Obviously I have to tell them to test themselves but right now I can calculate the probability that neither one of my parents have diabetes. What is this probability? Also calculate the probability that at least one has the disease. (Assume that the chances of one parent having diabetes is independent of the chances that my other parent suffers from it)

Explanation / Answer

Q5. (a) 8.3% poulation have diabetes in US; so P(diabetes) = 0.083

so Neither of the parents will have diabetes = P (dad no diabetes) * P( mom no diabetes) since both are independent events

P( Neither of them have diabetes) = ( 1- 0.083) * ( 1- 0.083) = 0.8408

P( only dad have diabetes) = P ( dad diabetes) * P ( mom no diabetes) = 0.083 * ( 1- 0.083) = 0.0761

P(bot have diabetes) = 0.083 * 0.083 = 0.00689

(b) Now it is clear that the child have diabetes so there are 3 possibilities from which child could have diabetes

1. the 4% chance when none of the parents have diabetes.

2. the 40% chance when both parents havediabetes

3. the 20% chance when one of the parents have diabetes.

soout of which, i guessed that none of my parents have diabetes so probability of that event

P( None of parents have diabetes but i have it) = ( 0.04 * 0.8408)/ (0.04 *0.8408 + 0.20 * 2 *0.0761 + 0.40 * 0.00689)

here this values are picked from the (a) part

P( None of parents have diabetes but i have it) = 0.033632/ 0.066828 = 0.5032= 50.32% or nearly half chances

P( atleastoneof them have disease and i retain it) =  (0.20 * 2 *0.0761)/ (0.04 *0.8408 + 0.20 * 2 *0.0761 + 0.40 * 0.00689)= 0.4554 = 45.54%

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