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I'm self-studying probability and I bumped into this question: suppose a physical trait (such as eye color) is based on one pair of genes and suppose that $d$ represents a dominant gene and $r$ a recessive gene. Thus a person with $dd$ is pure dominant, $rr$ is recessive, and $rd$ or $dr$ is hybrid. Children receive 1 gene from each parent and the occurrence of $r$ or $d$ in a gene is equally likely. If, with respect to a particular trait, 2 hybrid parents have a total of 4 children, what is the probability that 3 of the 4 children are pure dominant or hybrid in that trait?

I know how to do it: the probability of receiving $dd$ or $rd$ is $\frac{3}{4}$. By the probability mass function, you can simply do ${4\choose3}\left(\frac{3}{4}\right)^3 \frac{1}{4} = \frac{27}{64}.$

However, what's wrong with doing it this way:

The probability of receiving $dd$ is $\frac{1}{4}$. We first compute ${4\choose3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)$.

The probability of receiving $rd$ is $\frac{1}{2}$. We then compute ${4\choose3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)$.

Together, we have $$ {4\choose3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right) + {4\choose3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right) = \frac{19}{64}.$$

What did I miss?

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The first computation is correct; the second is incorrect, because it does not count those outcomes in which, of the three children with at least one dominant allele, there is at least one child who is homozygous dominant, and at least one child who is heterozygous dominant. That is to say, you have only counted those outcomes for which all three children are $dd$, or all three children are $rd$. You've not counted outcomes where you could have both $dd$ and $rd$ among the $3$ children.

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  • $\begingroup$ Very clearly! How would you compute the probability where you have both $dd$ and $rd$ among the 3 children? $\endgroup$ – user1691278 Sep 18 '18 at 4:24
  • $\begingroup$ @user1691278: Actually, the error is a bit more complicated than that. Since the last factor is the complement, you're not only missing cases, as described in the answer, but also counting cases you shouldn't count, namely dd dd dd rd and rd rd rd dd, where all four children have the dominant phenotype. $\endgroup$ – joriki Sep 18 '18 at 4:30

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