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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University officials hope that changes they have made have improved the retention rate. Last year, a sample of 20492049 freshmen showed that 1537 returned as sophomores. This year, 1516 of 1942 freshmen sampled returned as sophomores. Determine if there is sufficient evidence at the 0.05 level to say that the retention rate has improved. Let last year's freshmen be Population 1 and let this year's freshmen be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0Ha: p1=p2: Ha: p1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯p2 Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places. Step 3 of 3: Draw a conclusion and interpret the decision.
The coat color of black bears depends on a gene with two alleles. Allele B is dominant and confers black coat, whereas allele b is recessive and homozygous bears have white coats (albino black bears, I know it sounds weird but it is true). As we know from genetics each bear has two alleles and the coat colors are as follows: BB-black; Bb-black; bb-white. A researcher obtained the following data for the genotypes (BB, Bb or bb) of black bears: Genotype Number of bears BB 42 Bb 24 bb 21 You are performing the following hypothesis test: H0: The genotypes of the bears follow a binomial distribution, with a probability of receiving the b allele equal to the observed proportion of the b allele in the population. HA: The genotypes of the bears do not follow a binomial distribution What is the expected frequency of bears with genotype BB, Bb, and bb? Give the absolute expected frequency (number of bears) not the fraction/proportion of BB bears
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