An employee suspected of having used an illegal drug is given two tests that ope

Question

An employee suspected of having used an illegal drug is given two tests that operate independently of each other. Test A has probability 0.7 of being positive if the illegal drug has been used. Test B has probability 0.6 of being positive if the illegal drug has been used. What is the probability that neither test is positive if the illegal drug has been used? _______ All human blood can be "ABO-typed" as one of O, A, B, or AB, but the distribution of the types varies a bit among groups of people. Here is the distribution of blood types for a randomly chosen person n the United States. Choose a married couple at random. It is reasonable to assume that the blood types of husband and wife are independent and follow this distribution. What is the probability that the wife has type A blood and the husband has type B blood? 0.1 What is the probability that one member of the couple has type A blood and the other member of the couple has type B blood? .06

Explanation / Answer

1.

Test A: probability of test being positive ,P(A)=0.7

probability of test being negative, P(A-)=1-P(A)=1-0.7=0.3

Test B: Probability of test being positive, P(B)=0.6

probability of test being negative, P(B- )=1-P(B)=1-0.6=0.4

Probability of both test A and B negative, P(A- and B- )=P(A-).P(B-) by definition of condition independence

P(A- and B- )=P(A-).P(B-) = 0.3 *0.4 = 0.12

Probability of both test A and B negative is 0.12

2.

(i) Probability that the wife has type A blood and husband has type B blood:

as above given data probabbilty of blood group type applicable for all U.S. population irrespective of gender so

and blood type of husband and wife are indepedent

probability of wife has type A blood =P(A)=0.40

probability of husband have type B blood= P(B)= 0.12

by the rule of independence condition

probability of wife has type A blood and husband has type B blood , P (A and B)=P(A).P(B)=0.40*0.12=0.048

(ii) Probability that one member of couple has type A blood and the other member of couple has type B blood:

so there is two case

case I : wife has type A blood and husband has type B blood

as above (i) solution

P (wife has type A blood and husband has type B blood)=0.048

case II: wife has type B blood and husband has type A blood

as similar to (i) soltuion

P(wife has type B blood and husband has type A blood)=P(wife has B type blood).P(husband has A type blood)=0.12*0.40=0.048

and P(case I and Case II)= 0

so P( one member of couple has type A blood and the other member of couple has type B blood) = Pcase I or case II)= P(case I) +P(case II)- P(case I and Case II) = 0.048 +0.048 +0 = 0.096

Probability that one member of couple has type A blood and the other member of couple has type B blood is 0.096

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