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Answers without the blur. Sign up and see all textbooks for free! Q 104.

Expert-verified Found in: Page 333 ### The Practice of Statistics for AP

Book edition 4th
Author(s) David Moore,Daren Starnes,Dan Yates
Pages 809 pages
ISBN 9781319113339 # An athlete suspected of using steroids is given two tests that operate independently of each other. Test A has a probability $0.9$ of being positive if steroids have been used. Test B has a probability $0.8$ of being positiveif steroids have been used. What is the probability that neither test is positive if steroids have been used?$\left(a\right)0.72\left(c\right)0.02\left(e\right)0.08\left(b\right)0.38\left(d\right)0.28$

The correct option is $\left(c\right)0.02$

See the step by step solution

## Step 1. Given Information

$P\left(TestApositive\right)=0.9\phantom{\rule{0ex}{0ex}}P\left(TestBpositive\right)=0.8$

## Step 2. Concept used

Complement rule: $P\left({A}^{c}\right)=P\left(notA\right)=1-P\left(A\right)$

## Step 3. Calculation

The complement rule is $P\left({A}^{c}\right)=P\left(notA\right)=1-P\left(A\right)$

$P\left(TestAnegative\right)=1-P\left(TestApositive\right)=1-0.9=0.1\phantom{\rule{0ex}{0ex}}P\left(TestBnegative\right)=1-P\left(TestBpositive\right)=1-0.8=0.2$

According to the question, the two tests are independent of one another, hence the multiplication rule for independent events can be used:

Fill in the blanks with the corresponding probabilities:

$P\left(Neithertestpositive\right)=P\left(Bothtestnegative\right)\phantom{\rule{0ex}{0ex}}=P\left(TestAnegative\right)×P\left(TestBnegative\right)\phantom{\rule{0ex}{0ex}}=0.1×0.2\phantom{\rule{0ex}{0ex}}=0.02$

As a result, the proper option is $\left(c\right)0.02$ according to the option. ### Want to see more solutions like these? 