A population of 80 rats is tested for 4 genetic mutations after exposure to somechemicals: mutation A, mutation B, mutation C, and mutation D. 43 rats testedpositivefor mutation A. 37 rats tested positive for mutation B.39 rats tested positive formutation C. 35 rats tested positive for mutation D. One rat tested positive for all fourmutations. 5 rats tested positive for mutations A, B, and C. 4 rats tested positive formutations A, B, and D. 6 rats tested positive for mutations A, C, and D.3 rats testedpositive for mutations B, C, and D. 64 rats tested positive for mutations A or B. 63rats tested positive for mutations A or C. 59 rats tested positive for mutations A or D.58 rats tested positive for mutations B or C. 59 rats tested positive for mutations Bor D. 60 tested positive for mutations C or D. 8 rats did not show any evidence ofgenetic mutation.What is the probability that if 5 rats are selected at random, 3 will have exactly 2genetic mutations?Round your answer to five decimal places.

From the given information, A population of 80 rats is tested for 4 genetic mutations after…

Answered: A population of 80 rats is tested for 4…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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As mentioned in Experimental Question E11, red eyes is the wildtype phenotype. Several different genes (with each gene existing intwo or more alleles) are known to affect eye color. One allelecauses purple eyes, and a different allele causes sepia eyes. Both ofthese alleles are recessive to red eye color. When flies with purpleeyes were crossed to flies with sepia eyes, all of the F1 offspringhad red eyes. When the F1 offspring were allowed to mate witheach other, the following data were obtained:146 purple eyes151 sepia eyes50 purplish sepia eyes444 red eyesExplain this pattern of inheritance. Conduct a chi sq
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In a study of chromosomal anomalies observed in a randomly selected sample of 1200 infertile men with either a zero or low sperm count, a team of researchers assigned the value of 1 for the presence of any chromosomal anomaly in the subject's sperm and the value of 0 for the absence of all chromosomal anomalies in the subject's sperm.Of the 600 men with zero sperm count, 48 had chromosomal anomalies. Of the 600 men with low sperm count, 15 had chromosomal anomalies. The researchers would like to test the hypothesesHo: P1 = P2Ha: P1 does not equal P2where p₁ the true proportion of all men with zero sperm count that have chromosomal anomalies and p₂ = the true proportion of all men with low sperm count that have chromosomal anomalies.What is the z standardized test statistic, for this test?
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