BIT review - exclusive! 14. P(AorB)= P(A) + P(B) <1 if...

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1. Mutually Exclusive: No points in common 2. Probability of B: No signal = 175 Not all channels working = 140 Bad Reception = 85 (175+140+85)/500= .80 3&4. A and B have Bad Reception in common so NOT mutually exclusive Upside Down U = and U = or | = given 5. P(AandB) = 85/500= .12 6. P(B|A)= P(AandB)/P(A) P(A)= (85+55)/500= .28 .17/.28 = .61 P(B|A)= P(B) means Independent P(A|B)= P(A) means Independent but .61 is NOT .80 so NOT independent P(A|B) = P(AandB) = 0 means Mutually Exclusive 8. P(A or B)Complement= P(A) + P(B) – P(AandB) .28 + .80 - .17 = .91 1-.91= .09 9. .45 10. 35/75 = .47 P(F|R)= P(FandR)/P(R) = (35/200)/(75/200)= 35/75 11. EASY 1-.4 = .6 12. P(AandB”complement”)= P(A)-P(AandB) P(AorB)= P(A) + P(B) – P(AandB)= .8= .11 – x= *DRAW OUT GRAPHS TO HELP YOU OUT!!!!!!!!!!! 13. Because if mutually exclusive P(B|A) = P(AandB)/P(A) should be equal to 0 because Mutually
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Unformatted text preview: exclusive! 14. P(AorB)= P(A) + P(B) <1 if P(A) = .6 P(B) < 1-.6 = .4 15. EASY 16. P(A)= .8 P(B)=.65 P(AorB)=.78 P(B|A)= P(BandA)/P(A) P(AorB)= P(A) + P(B) P(AandB)= .78= .8 + .65 x = .8375 17. P(A|B)= P(A) in order to be independent P(AandB)/(PB)= .3/.6 = .5 = P(A) = YES they are independent 18. 19. .5 so NONE of the above bc half chance for both. 20. 21. E(x) = E * F(x) X= 1, 2, 3 F(x) = 1/6, 2/6, 3/6 E(x) = 1(1/6) + 2(2/6) + 3(3/6) = 2.333 22. 23. P=.5 n*p = 50 n= 100 24. P=.4 n=50 : np(1-p)= 50(.4)(.6)=12 25. C (8 on top) (2 on bottom) = .6^2 * .4^6 8!/6!2! = .28 26. C(7 8) = .6^7 * .4^1 + C(8 8) = .6^8 * .4^0 C(8 7)= 8 C(8 8)=1 .4^0=1 8 (.6^7)(.4^1)+(.6^8)...
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This note was uploaded on 10/26/2009 for the course BIT 2405 taught by Professor Plkitchin during the Spring '08 term at Virginia Tech.

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BIT review - exclusive! 14. P(AorB)= P(A) + P(B) <1 if...

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