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anonymous

  • one year ago

HELP PLEASE!! Calculate probability of the independent event P(A or B) when P(a)=.3 and P(B)=.7

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  1. anonymous
    • one year ago
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    You just have to multiply .3 by .7

  2. anonymous
    • one year ago
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    oh ok well what about conditional probability with the same variables?

  3. ybarrap
    • one year ago
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    $$ P(A\cup B)=P(A)+P(B)-P(A\cap B) $$ Since A and B are independent, \(P(A\cap B)=0\) So just SUM the independent probabilities.

  4. anonymous
    • one year ago
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    so just 1?

  5. ybarrap
    • one year ago
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    Yes. See - http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle

  6. ybarrap
    • one year ago
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    Specifically, this section - http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#In_probability

  7. anonymous
    • one year ago
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    ok what about if the variables are P(a)=.5 P(b)=.4 and then P(a and b)=.1?????

  8. anonymous
    • one year ago
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    I'm supposed to calculate the probability

  9. ybarrap
    • one year ago
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    You've got everything you need, P(a), P(b) and P(a and b) -- see my formula above! Just plug in for P(a or b).

  10. ybarrap
    • one year ago
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    $$ P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.5+0.4-0.1 $$ See how I did that?

  11. anonymous
    • one year ago
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    yes i just thought a and b were supposed to = P(A and B) but the answer is .8

  12. ybarrap
    • one year ago
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    Are you asking me?

  13. anonymous
    • one year ago
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    yes(:

  14. ybarrap
    • one year ago
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    What are you asking, not sure

  15. anonymous
    • one year ago
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    the answer is .8 right?

  16. ybarrap
    • one year ago
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    Yes for P(a or b).

  17. anonymous
    • one year ago
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    ok but theres nothing else i need to find? it just says calculate the probability

  18. ybarrap
    • one year ago
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    Tell me the whole question to be sure

  19. anonymous
    • one year ago
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    its 27

  20. ybarrap
    • one year ago
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    Oh, those are different problems that P(A or B) Ok

  21. ybarrap
    • one year ago
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    Do you know what P(A|B) means?

  22. anonymous
    • one year ago
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    i know its a conditional probability problem

  23. ybarrap
    • one year ago
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    $$ P(A|B)=\cfrac{PA\cap B}{P(B)}=\cfrac{0.1}{0.4} $$ You can do the next. BRB

  24. anonymous
    • one year ago
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    i got .25

  25. ybarrap
    • one year ago
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    Do #27 now

  26. anonymous
    • one year ago
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    umm isn't that what i did?

  27. anonymous
    • one year ago
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    oh okay for 27 i got 4

  28. anonymous
    • one year ago
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    if you can check the other ones i got .21 on 23 and on 25 i got 1 but if you can't then thank you for helping!

  29. ybarrap
    • one year ago
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    Probabilities can not be greater than 1 so #27 is not 4

  30. ybarrap
    • one year ago
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    For #27 what do you have for P(B|A)=P(B and A)/P(A) ?

  31. ybarrap
    • one year ago
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    We already did #23

  32. anonymous
    • one year ago
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    oh well you had .1/.4 so in 27 i just did .4/.1

  33. ybarrap
    • one year ago
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    No. The "|" does NOT mean division. It means "given" P(A|B) means Probability of event A given event B.

  34. anonymous
    • one year ago
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    you did .1/.4

  35. ybarrap
    • one year ago
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    $$ P(B|A)=\cfrac{P(A\cap B)}{P(A)}\\ P(A|B)=\cfrac{P(A\cap B)}{P(B)} $$ Do you see the difference above between the two?

  36. ybarrap
    • one year ago
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    Yes and that one is correct

  37. anonymous
    • one year ago
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    i guess I'm not understanding this then

  38. ybarrap
    • one year ago
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    So we are dealing with probabilities of events. When we are given that an event occurs, we are asked what is the probability that the other will occur. If they are independent, then there is no effect. But if they are not independent, then conditional probability tells us that Probability of A given B is the Probability of A and B divided by the probability of B. Here is a visual |dw:1433295585568:dw|

  39. anonymous
    • one year ago
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    ok

  40. ybarrap
    • one year ago
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    That fraction is $$ P(A|B)=\cfrac{P(A\cap B)}{P(B)} $$

  41. anonymous
    • one year ago
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    that looks like alien writing to me

  42. ybarrap
    • one year ago
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    $$ P(A|B)=P(A \text{ given }B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{P(A\text{ and }B)}{P(B)} $$ Is this better?

  43. anonymous
    • one year ago
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    haha no I'm sorry... what does it look like with the values plugged in

  44. ybarrap
    • one year ago
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    |dw:1433296024303:dw|

  45. ybarrap
    • one year ago
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    $$ P(A|B)=P(A \text{ given }B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{P(A\text{ and }B)}{P(B)}=.1/.4 $$ For #26

  46. ybarrap
    • one year ago
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    For #27 $$ P(B|A)=P(B \text{ given }A)=\cfrac{P(A\cap B)}{P(A)}=\cfrac{P(A\text{ and }B)}{P(A)}=.1/.5 $$

  47. anonymous
    • one year ago
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    ok so its .2

  48. anonymous
    • one year ago
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    it just made a lot more sense

  49. anonymous
    • one year ago
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    so because (a and b) the value is .1 and P(a)=.5

  50. ybarrap
    • one year ago
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    #24 $$ P(A|B)=P(A \text{ given }B)=\cfrac{P(A\cap B)}{P(B)}=\cfrac{P(A\text{ and }B)}{P(B)}=(0.3\times0.7)/0.7 $$ Notice that the result is independent of B because P(A|B) = P(A)

  51. ybarrap
    • one year ago
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    This happened because A and B are Independent

  52. ybarrap
    • one year ago
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    If you understood that, then you can do #25 without any calculations

  53. ybarrap
    • one year ago
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    Correct on your last comment

  54. anonymous
    • one year ago
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    ok thank you i got it now

  55. ybarrap
    • one year ago
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    You're welcome

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