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Homework Help: Probability - 2 events HELP

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data

    The two events A and B have probabilities 0.2 and 0.4. Also P (A n B)=0:08.
    (a)Are the two events A and B independent? Explain.
    (b) Find the probability that either A or B or both occur.
    (c) Find the probability that neither A nor B occurs.
    (d) Find the probability that exactly one of A or B occurs.


    2. Relevant equations
    P(A n B) = P(A)*P(B)

    3. The attempt at a solution

    My working out:

    (a) For 2 events to be independent they must satisfy:
    P(A n B) = P(A)*P(B)
    so 0.08 = 0.2x0.4
    which makes them independent .. Correct ?

    (b) hmm not sure about this one! need some helpp

    (c) P(neither A or B) = 0.8*0.6 Correct?

    (d) hmm maybe P = 0.2*0.6 + 0.4*0.8 = 0.44 ?????
     
  2. jcsd
  3. Aug 31, 2010 #2

    eumyang

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    Homework Helper

    (A) is right.

    For (B), you need a formula for [tex]P(A \cup B)[/tex]. But note that the problem ends with "or both."

    For (C), what would be the compliment of "neither A nor B"?

    (D) is wrong. You again need to find [tex]P(A \cup B)[/tex], but note the difference between (D) and (B).


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  4. Sep 1, 2010 #3
    so for (B) i should use: [tex]P(A \cup B)[/tex] = P(A) + P(B) - P(A n B)

    for (C) use P(A^c n B^c) = 1 - [tex]P(A \cup B)[/tex] ??

    for (D) use: [tex]P(A \cup B)[/tex] = P(A) + P(B)

    Need help asap

    thankss
     
  5. Sep 1, 2010 #4

    eumyang

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    Homework Helper

    You sure about these? Again, the end of (B) says "or both". [tex]P(A \cap B)[/tex] would represent the "both," would it not? And in (D), we want "exactly one," so we cannot include both A and B.


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  6. Sep 1, 2010 #5
    ok so would it look something like that

    so for (B) i should use: [tex] P(A \cup B)[/tex] = P(A) + P(B) + P(A n B)

    for (D) use: [tex]P(A \cup B)[/tex] = P(A) + P(B) - P(A n B)
     
  7. Sep 1, 2010 #6

    HallsofIvy

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    Science Advisor

    For (B), he is correct. "[itex]P(A\cap B)[/itex]" is already included in both P(A) and
    P(B). In order to count it only once we must subtract off one: [itex]P(A\cup B)= P(A)+ P(B)- P(A\cap B)[/itex].

    So for D, you subtract off [itex]P(A\cap B)[/itex] completely- twice.
     
  8. Sep 1, 2010 #7
    what do u mean by subract off [itex]P(A\cap B)[/itex] completely ?
    so - [itex]P(A\cap B)[/itex] twice ?
     
  9. Sep 1, 2010 #8

    eumyang

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    Homework Helper

    That's it, no more posting early in the morning. >.<


    69
     
  10. Sep 3, 2010 #9
    Thanks for the help guys
    Appreciate it
     
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