Posted by:
BoaMorph
at Mon Jun 11 22:41:56 2007 [ Email Message ] [ Show All Posts by BoaMorph ]
as follows:
For the two 66% possible het parents, if we knew nothing about their offspring, the probability of both being het blood is:
(2/3)*(2/3) = 4/9 = 44.4%
The probability of only one of them being het blood is:
(2/3 * 1/3)+(1/3 * 2/3) = 4/9 = 44.4%
The probability of neither being het blood is:
(1/3)*(1/3) = 1/9 = 11.1%
Now if we consider your litter, there were 11 offspring, all normal phenotype (no bloods). The probability of this occuring is:
1-(0.75^11) = 0.042 = 4.2%
What does this mean? Based on your litter the probability that at least one of the parents is not het blood is 95.8%. The good news is, that there is still a 4.2% probability that both parents are het blood - much better than your odds of winning the lottery - so breed them again next year.
The even better news is that even if you assume the worst case and figure that based on your litter it is much more likely (95.8%) that at least one of the parents is not het blood, you still have a 66% probability of the other parent being het blood.
If you were to get another similar sized litter next year with no bloods, the probability becomes extremely high that at least one of the parents is not het blood (another litter of 11 normals results in a probability of 1-0.75^22 = 99.8% that at least one parent is not het blood). Even then, with it being a near certainty that one parent is not het blood, there is still a 66% probability of the other parent being het blood - breeding these two together will yield no further information about whether one or zero of the pair are het blood.
At that point, to determine whether one or neither of these boas is het blood (and if one is het blood, to determine which one), you would have to breed each of them to blood or 100% het blood boas and see what you get.
Hope this helps! Best regards,
Steve Reiners
www.BoaMorph.com
[ Hide Replies ]
|
The probabilities are..... - BoaMorph, Mon Jun 11 22:41:56 2007
