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anonymous

  • 5 years ago

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.).

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  1. anonymous
    • 5 years ago
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    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?

  2. anonymous
    • 5 years ago
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    it would be easier to use a punnett square

  3. anonymous
    • 5 years ago
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    Quick hint: intuition should tell you the chance they both have the disease should be smaller than the chance either alone has it.

  4. anonymous
    • 5 years ago
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    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)

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

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