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The probability that a child in a given family will inherit a certain genetic mutation is 5%, independently of their siblings. Consider a family with 4 children.

(a) If it is known that the first born has inherited the mutation, what is the expected number of children in the family who have inherited the mutation?

(b) If it is known that at least one child has inherited the mutation, what is the expected number of children in the family who have inherited the mutation?

For part a), I got the answer which is 1 + 0.05 * 3 = 1.15. However, this is the same answer I get for part b), when the answer is supposedly 1.08. Can someone please suggest a solution for b)?

Anon
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  • A priori the number of children with the mutation is one of ${0,1,2,3,4}$. You can easily work out the probabilities of each. Being told that at least one child has it excludes $0$, so you can now renormalize the probabilities and read off the expected value. – lulu Aug 30 '23 at 18:48
  • Could you please show me the calculation? Why does applying the Law of Total Probability to find probabilities of there being 1,2,3,4 children and then summing for the expected value not work? – Anon Aug 30 '23 at 20:57
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    If the a priori probabilities are ${p_0, p_1, p_2,p_3,p_4}$ then, once $0$ is excluded, the new probabilities, for ${1,2,3,4}$ are $p_i'=\frac {p_i}{1-p_0}$. Now the expectation is $\sum_{i=1}^4ip_i'$. This is all easy to compute and yields expectation $=1.078203444$. – lulu Aug 30 '23 at 21:03
  • should say, this is all a variant of the usual Boy-Girl probability "paradox". That is to say, it's very different to be told that "at least one of two children is a girl" or "the eldest of the two children is a girl." – lulu Aug 30 '23 at 21:06

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