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Intro Statistics/ Probability help?

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    1) A computer firm presently has bids out on three projects. Let Ai = {awarded project i} for i = 1,2,3.
    Suppose that P(A1) = 0.20, P(A2) = 0.25, P(A3) = 0.28, P(A1∩A2) = 0.10, P(A1∩A3) = 0.06, P(A2∩A3) =
    0.08, and P(A1∩A2∩A3) = 0.01.
    Compute the following probabilities: Hint: Draw a Venn diagram

    A) Prob of (A2Compliment 'intersection' A3Compliment)

    I'm confused with this one. I did draw a Venn diagram.
    Am I supposed to say...

    A2compliment * A3 compliment= 0.75*0.72=0.54.
    Or compute it as 1- P( the union of A2 with A3)= 1-[0.25+0.28-0.08]=0.55


    The computers of seven faculty members in a certain department are to be replaced. Three of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the seven.

    a. What is the probability that both selected setups are for laptop machines?

    What I did was:

    A= 1st is laptop
    B= 2nd is laptop
    P(A intersection B)= (3/7) * (2/6) = 1/7

    b. What is the probability that at least one computer of each type is chosen for setup?
    This one confuses me because I know he said "at least one" means the union of two events, but the question says "at least one computer of each type is chosen"... and since the problem states that two computers will be chosen each day, is is still the union of two events?

    What I did:


    1) P[ (A intersect B) union with (B intersect A) ]= (3/7)*(4/7) + (4/7)*(3/7)= 24/49

    then I did it this way

    2) Prob. ot at least one computer of each type chosen=
    1 - [ (A intersect A) + (B intersect B) ] =1 - [ (3/7)(3/7) + (4/7)(4/7)] = 24/49


    The proportions of blood phenotypes in the U.S. population are as follows:
    A= 0.40, B= 0.11, AB= 0.04, O=0.45

    Assuming that the phenotypes of two randomly selected individuals are independent of one another,

    A. what is the probability that both phenotypes are O?

    what I did: 0.45*0.45= 0.20

    B. What is the probability that the phenotypes of two randomly selected individuals match?
    What I did:

    Added the probabilities of
    (A intersect A) + (B intersect B) + (AB intersect AB) + (O intersect O)=

    0.40*0.40 + 0.11*0.11 + 0.04*0.04 + 0.45*0.45= 0.38


    Consider the system of components in the accompanying picture. Components 1 & 2 and 3 & 4 are connected in series (call these subsystems 12 & 34). A subsystem will work only if both components work. In order for the entire system to function, it must be the case that at least one subsystem works.

    Suppose that each individual component functions independently of all other components and each component functions with probability 0.8.

    A. What is the probability that subsystem 12 does not work or P( A1Compliment 'union' A2Compliment)?

    What I did:

    0.2 +0.2= 0.40

    B. Find the probability that the entire system functions or P( A12 'union' A34)?
    What I did:

    1- P( A12Compliment 'intersection' A34Compliment)=

    1- [ P (A1Compliment 'union' A2Compliment) 'intersect' P ( A3Compliment 'union'
    A4Compliment)] = 1- [ 0.40 * 0.40] = 1- 0.16= 0.84

    I don't know what I'm doing....Thanks for your assistance!
  2. jcsd
  3. Feb 7, 2012 #2


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    The second method is correct. The first method is incorrect unless A and B are independent.


    You have to be more careful with your events. "A = laptop" and "B = desktop" are ambiguous, because the probabilities of these events will change depending on which combination of events is chosen.

    Try it this way:

    A = 1st is laptop
    complement of A = 1st is desktop

    B = 2nd is laptop
    complement of B = 2nd is desktop

    Then you could calculate using either of your methods:

    [tex]P[(A \cap B^c) \cup (B \cap A^c)][/tex]
    [tex]1 - P[(A \cap B) \cup (A^c \cap B^c)][/tex]
  4. Feb 7, 2012 #3


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    No, you can't just add the probabilities unless the events are mutually exclusive, i.e. unless it's impossible for both components to fail. In general, you need to use:

    [tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]

    Correct up to this point, but this part is wrong:

    for the same reason I noted above.
  5. Feb 7, 2012 #4

    Thanks a lot for your help.

    I'll correct it.
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