problemset4solution - PAM 2100 Spring 2009 Problem Set #4...

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PAM 2100 Spring 2009 Problem Set #4 Answer key 10 points From the textbook, solve the following problems: (Correct final answers get full credit. Showing work is not necessary) 4.21 a) P(type O blood)=1- (.4+.11+.04)=.45 (.5 points) b) It is the probability that the randomly chosen person is type O or type B. So, P( type O blood or type B blood)=P(type O blood)+P(type B blood) =.45+.11=.56 (.5 points) 4.22 Prob (Both are type O) =(.45)(.35)= .1575 (.5 points) Prob (both are the same type) =P (both are type O or both are type A or both are type B or both are type AB ) =P(both are type O)+ P(both are type A)+ P(both are type B)+ P(both are type AB)= (.45)(.35)+ (.40)(.27)+ (.11)(.26)+ (.04)(.12)= .2989 (1 point) 4.34 (Note that since individuals appeared randomly, the event that any particular individual is a universal donor is independent of other individuals being universal donors. So we can apply the multiplication rule for independent events) Prob (a single individual is not O-negative)=(1-0.07)=.93
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This note was uploaded on 05/09/2009 for the course PAM 2100 taught by Professor Abdus,s. during the Spring '08 term at Cornell University (Engineering School).

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problemset4solution - PAM 2100 Spring 2009 Problem Set #4...

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