anonymous
  • anonymous
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.).
Mathematics
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anonymous
  • anonymous
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.).
Mathematics
katieb
  • katieb
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
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
  • anonymous
it would be easier to use a punnett square
anonymous
  • anonymous
Quick hint: intuition should tell you the chance they both have the disease should be smaller than the chance either alone has it.

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

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