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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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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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