Practice Midterm 2 WI 10 Solutions

# Note that pnot a 1 pa 1 4 6 pat least 1 a 1 pno as 1

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Note that P(not A) = 1 – P(A) = 1 - .4 = .6 P(At least 1 A) = 1 – P(no A’s) = 1 – P(not A)^ 2 = 1 – (.6)^2 = 0.64 OR P(At least 1 A) = P(AA or AB or AC or BA or CA) = P(AA) + P(AB) + P(AC) + P(BA) + P(CA) = (0.4)(0.4) + (0.4)(0.35) + (0.4)(0.25) + (0.35)(0.4) + (0.25)(0.4) = 0.64 ANS____ 0.64 ____ 30. What is the chance that neither one got an A? P(Neither has an A) = 1 – P(At least 1 A) = 1 – 0.64 = 0.36 ANS__ 0.36 ________ 7

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ANSWER SHEET MC AND TF SOLUTIONS Stat 133 Midterm 2 WI 2010 THESE REPRESENT YOUR FINAL ANSWERS! BLANKS WILL RECEIVE 0 POINTS. 1. A B C D 2. A B C D 3. A B C D 4. A B C D 5. A B C D 6. A B C D 7. A B C D 8. A B C D 9. A B C D 10. A B C D 11. A B C D 12. A B C D 13. A B C D 14. A B C D 15. A B C D 16. A B C D 17. A B C D 18. A B C D 19. A B C D 20. A B C D 8
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