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e Do Homework CRYSTAL ROSSI Mozilla Firefox m/Student/PlayerH 4&flushed; false&c

ID: 3224540 • Letter: E

Question

e Do Homework CRYSTAL ROSSI Mozilla Firefox m/Student/PlayerH 4&flushed; false&cld; 4358479&centerwin; yes https://www. mathxl.col STA2023 STATISTICS ON LINE RUSTEM MULYUKOV 564438 Homework: Section 4-4 HW Part A 2 of 9 (9 complete) Refer to the table below. Given that 2 of the 126 subjects are randomly selected, complete parts (a) and (b) Grou Rh 36 Type 33 Rh a. Assume that the selections are made with replacement What is the probability thatthe 2 selected subjects are both group Band type Rh (Round to four decimal places as needed.) Enter your answer in the answer box and then click Check Answer. 1 part Clear All remaining a fr Search the web and Windows 90% M. CRYSTAL ROSE 4/21/17 56 AM Save Assigned Media EQuestion Hel Check Answer 1:56 AM OneDrive Desktop A Ee 4/21/2017

Explanation / Answer

a. Total Number of subjects = 126

Number of subjects with Group B and type Rh+ = 16

Assume selections are made with replacement.

Probability that selected subjects are both group B and and type Rh+

Probability of first subject with Group B and type Rh+ = Number of subjects with Group B and type Rh+ / Total Number of subjects = 16/126

As selections are made with replacement for selecting second person there is no change in the  Number of subjects with Group B and type Rh+ and Total Number of subjects

Probability of seccond subject with GroupB and type Rh+ = Number of subjects with GroupB and type Rh+ / Total Number of subjects = 16/126

Probability that selected subjects are both group B and and type Rh+ = Probability of first subject with GroupB and type Rh+ x Probability of seccond subject with GroupB and type Rh+ = (16/126)(16/126) =0.0161

b. Total Number of subjects = 126

Number of subjects with Group AB and type Rh+ = 16

Assume selections are made without replacement.

Probability that selected subjects are both group AB and and type Rh+

Probability of first subject with GroupAB and type Rh+ = Number of subjects with GroupAB and type Rh+ / Total Number of subjects = 16/126

As selections are made without replacement for selecting second person, Number of subjects with GroupAB and type Rh+ is reduced by =15 and Total Number of subjects also reduced by 1 = 126-1=125

Probability of seccond subject with GroupAB and type Rh+ = Number of subjects with GroupAB and type Rh+ / Total Number of subjects = 15/125

Probability that selected subjects are both group AB and and type Rh+ = Probability of first subject with GroupAB and type Rh+ x Probability of seccond subject with GroupAB and type Rh+ = (16/126)(15/125) =0.0152