MAT 2379 Quiz: Assignment 2 Solution

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maroonrabbit19

15 Nov 2021

School

University of Ottawa

Department

Mathematics

Course

MAT 2379

Professor

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Document Summary

Only questions 1, 3, 5 and 7 will be marked. [10] blood type o and 12 donors who do not have blood type o. (a) for x = 0, we must determine the probability that neither one of the donors has blood type o. Suppose that we select the rst donor and we observe that the sample does not have blood type o. Thus, the probability that the rst one does not have blood type o is 12/22. In the remaining 21 donors, there are 10 donors with blood type o and 11 donors who do not have blood type o. This means that in the remaining 21 donors, there are. 10 donors with blood type o and 11 donors who do not have blood type o. So the conditional probability that the second donor does not have blood type o given that the rst one does not have blood type o is 11/21.

Related Questions

Hardy-Weinberg Law

Common blood types are determined genetically by three alleles A, B, and O. (An allele is any of a group of possible mutational forms of a gene.) A person whose blood type is AA, BB, or OO is homozygous. A person whose blood type is AB, AO, or BO is heterozygous. The Hardy-Weinberg Law states that the proportion of heterozygous individuals in any given population is

where represents the percent of allele A in the population, represents the percent of allele B in the population, and represents the percent of allele O in the population. Use the fact that