Records show that a child of parents with heart disease has a probability of 5/6 of inheriting the disease. Assuming independence, what is the probability that, for a couple with heart disease that have two children:
(a) Both children have heart disease:
(b) Exactly one child had heart disease
Answers should be exact (a common fraction as an answer is allowed.).

Question

Records show that a child of parents with heart disease has a probability of  5/6 of inheriting the disease. Assuming independence, what is the probability that, for a couple with heart disease that have two children: 
(a)	Both children have heart disease: 
(b)	Exactly one child had heart disease
Answers should be exact (a common fraction as an answer is allowed.).

anonymous · Answer

Since they are independent, there is a $\frac{5}{6}$ chance the first child will have the disease, and then the same chance the second one will have it. What's the chance that both of these will occur together?

anonymous · Answer

it would be easier to use a punnett square

anonymous · Answer

Quick hint: intuition should tell you the chance they both have the disease should be smaller than the chance either alone has it.

anonymous · Answer

a.

P(Both have heart disease) = (5/6)(5/6)

b.  P(Exactly one has heart disease) = P(A and not B) + P(Not A and B) = (5/6)(1/6) + (1/6)(5/6)

anonymous · Answer

there was one part of the question i forget 

c. neither child have diesease