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Q 104.

Expert-verifiedFound in: Page 333

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 positive

if 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 $\left(c\right)0.02$

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

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

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

Use the complement rule to help you:

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

$P(Neithertestpositive)=P(Bothtestnegative)\phantom{\rule{0ex}{0ex}}=P(TestAnegative)\times P(TestBnegative)\phantom{\rule{0ex}{0ex}}=0.1\times 0.2\phantom{\rule{0ex}{0ex}}=0.02$As a result, the proper option is $\left(c\right)0.02$ according to the option.

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