# Genetics and probability (“other than” case)

(Moved from Bio SE due to the mathematical nature of the problem.)

If a man and woman, both carriers of a autosomal recessive disorder (i.e. having genotype $Aa$), produce three children, what is the probability of one or more children having the disorder?

The answer key suggests finding the probability of the children all being normal and then subtracting that from 1: $$Aa \times Aa$$ $$P_{AA\:or\:Aa}=\frac{3}{4}$$ $$\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{27}{64}\tag{3 normal}$$ $$1-\frac{27}{64}=\frac{37}{64}\tag{≥1 abnormal}$$ I understand the logic behind that: If they aren’t all normal, then at least one of them has to be abnormal.

However, what if I didn’t want to use the $1-x$ method? What if I wanted to do it the hard way?

I tried breaking it into three cases: 1 abnormal, 2 abnormal, and all abnormal. $$P_{aa}=\frac{1}{4}$$ $$\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{64}\tag{1 abnormal, 2 normal}$$ $$\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{64}\tag{2 abnormal, 1 normal}$$ $$\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\tag{3 abnormal, 0 normal}$$ $$\sum P=\frac{13}{64}$$ which obviously isn’t the correct answer. What’s wrong with the way I’m doing it?

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You have to take in accounte repetitions. There are three cases with one abnormal and two normal (namely: the first is abnormal, the second is abnormal, the third is abnormal). Similarly with 2 abnormal and 1 normal. So you have to multiply those probabilities thimes 3 and then add up. For general curiosity: you are in fact using "binomial distribution". – Cornelis Feb 21 at 9:54

$$P_{aa}=\frac{1}{4}$$ $$\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{64}\tag{1 abnormal, 2 normal}$$ $$\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{64}\tag{2 abnormal, 1 normal}$$ $$\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\tag{3 abnormal, 0 normal}$$ The error in your argument lies in the first and second statements. Suppose the three children are $A,B,C$.

In the first case, where 1 is abnormal and 2 are normal, we can have,

Abnormal - $A$, Normal - $B,C$

Abnormal - $B$, Normal - $C,A$

Abnormal - $C$, Normal - $A,B$

Thus, this condition is satisfied in three outcomes, while only one has been counted. The probability for this case then becomes, $$3\times\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{27}{64}$$ Similarly, in the second case,

$$3\times\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{9}{64}$$ which gives the desired result.

In general, to get the number of outcomes for the particular case, we need to find out under how many combinations that case can occur. In the first case here, 1 child is abnormal, and can be selected from 3 children in $\binom{3}{1}=3$ ways. Thus, we have three such outcomes possible.

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Thank you. I especially liked your inclusion of the combination notation. – lightweaver Feb 21 at 10:17

Your mistake is that you specify that the first will be abnormal in the calculation $$\frac14\times\frac34\times\frac34$$ To obtain the correct result you should sum up this probability three times instead, as any one of the three children can be abnormal. This gives you $$3\times \frac14\times\frac34\times\frac34$$ or equivalently (to make it clear)$$\left(\frac14\times\frac34\times\frac34\right)+\left(\frac34\times\frac14\times\frac34\right)+\left(\frac34\times\frac34\times\frac14\right)=3\times \frac14\times\frac34\times\frac34$$ to be the probability of exactly one child to be abnormal. The same for two abnormal children. Of, course if all three are normal or abnormal then there is only one such way. This gives the result $$3\times \frac{9}{64}+3\times \frac{3}{64}+\frac1{64}=\frac{37}{64}$$ which is the correct result (comparing the other method).

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To begin, let's flip some coins!

The probability of flipping a fair coin one time and scoring a heads is

$1 - \frac{1}{2} = 0.5$

because there are two possible outcomes (heads [H] or tails [T]). It is important to note that these are independent events. If you flip the coin twice you have four possible scenarios; HH, TH, HT, TT. Three of these scenarios features a head. Therefore the probability of heads occurring once in two flips is

$1 - \frac{1}{2}^2 = 0.75$

The probability of two heads occurring from two flips is

$\frac{1}{2}^2 = 0.25$

Three flips gives eight outcomes (HHH, HHT, HTH, THH, HTT, THT, TTH, TTT), of which, seven give at least one head. Therefore the probability of scoring at least one head in three coin flips is

$1 - \frac{1}{2}^3 = 0.875$

You should be able to see a common pattern forming, the calculation for the probability of a single outcome occurring in a given number of events is:

$1 - \frac{X}{Y}^n$

where $Y$ is the number of possible outcomes per event (2, heads or tails), $X$ is the number of outcomes other than your focal ones (1, tails), and $n$ is the number of independent events (number of times the coin is flipped). So the probability of flipping a single head in 10 coin flips is

$1 - \frac{1}{2}^{10} = 0.99$

Now you can apply this to your genetics question. Each parent passes one allele to their offspring, because alleles are passed randomly, so the probability of receiving an $a$ from the father is

$1 - \frac{1}{2} = 0.5$

and the same for receiving an $a$ from the mother is the same. There are four possible outcomes in the offspring (AA, Aa, aA, aa). The probability of being $aa$ is (the same as scoring two heads from two coin flips)

$\frac{1}{4} = 0.25$

The genotype of each child is determined by two independent events each with a probability of 0.5, and for each child the probability of being $aa$ is 0.25. The probability that at least one of three children is $aa$ can be calculated. You have three children, these is the number of events ($n$), each will have one of four possible genotypes ($Y$), and three of these are not $aa$ ($X$). Thus the probability that one or more of the children is $aa$ is

$1 - \frac{X}{Y}^n$

$1 - \frac{3}{4}^3 = 0.578$

As proof, the probability that the parents have one child and they are $aa$ is

$1 - \frac{3}{4}^1 = 0.25$

The result is 0.578 (37/64) is because there are 64 possible outcomes (AA-AA-AA, AA-AA-Aa... aa-aa-aA, aa-aa-aa), of which, 37 feature at least one child with $aa$. This is analogous to the classic genetic code table where there are three nucleotides in the codon, and four nucleotides possible at each position in the codon. For example, 37 have a G somewhere in the codon:

I started this answer on Biology SE where the question was originally posted, but it was deleted by the time I went to post!

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