Subject: Risk Assessment - part 2

To continue I must introduce the concept of conditional probability. If we randomly draw a dog from the proposed general population pool and have no prior information about it, we expect (given our hypothetical rate of affecteds equal to 1% of the population) that, on average, the dog will be affected 1 time in 100; it will be clear 81 times in 100; and it will be a carrier 18 times in 100.

However, if we randomly draw a dog from the population pool but exclude all affected dogs, i.e., select only unaffected dogs, then:

An "unaffected" dog will be clear 81/(81 + 18) = 0.81818 or 81.8 times in 100; and it will be a carrier 18/(81 + 18) = 0.18181 or 18.2 times in 100. The fact that we have excluded affecteds "conditions" the result.

Similarly, if a dog has produced an affected pup, then we know that the dog must either be a carrier or it must be affected. The probability that a "produced" dog is a carrier is 18/(1 + 18) = 0.9474 or 94.7 times in 100; and the probability that it is affected is 1/(1 + 18) = 0.0526 or 5.3 times in 100.

Observe that the conditioning factor which appears in the denominator in these calculations is simply the total fraction of the dogs remaining in the pool from which our selection is made. You should also note that if a dog has been designated as either affected, carrier, or clear, then no conditioning calculations are required, since, in effect, we claim to know the genotypic character of the dog and need make no further adjustments. For example, if a dog has produced an affected pup, but is itself not affected, then its proper classification is "carrier" rather than either "produced" or "unaffected."

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Jim Seltzer
Willowind Dalmatians
http://users.nbn.net/~jseltzer
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