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Math Help - workers probability

  1. #1
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    workers probability

    In a large factory with many workers , 30 % of the workers belong to union A , 60 % Union B and the remaining 10 % doesn't join any union . A sample of 5 workers is selected .

    (1) Find the probability that the sample has less than 2 members of UNion A .

    (2) It's found that exactly 4 of these workers are members of union . What is the probability that more of the members come from Union A than UNion B .

    so For (1) ,

    it could be 1 member from A , then 4 from A' OR 0 from A and 5 from A'

    P(x<2)=(0.3)(0.7)^4+(0.7)^5=0.2401

    (2) I did not consider the one member who has no union because it doesn't make a difference so Union A - 4 , B - 0 , OR A-3 , B-1

    P(A>B)=(0.3)^4+(0.3)^3(0.6)=0.0243

    But unfortunately both are wrong . The answer given is (1) 0.528 (2) 1/9
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    In a large factory with many workers , 30 % of the workers belong to union A , 60 % Union B and the remaining 10 % doesn't join any union . A sample of 5 workers is selected .

    (1) Find the probability that the sample has less than 2 members of UNion A .

    (2) It's found that exactly 4 of these workers are members of union . What is the probability that more of the members come from Union A than UNion B .

    so For (1) ,

    it could be 1 member from A , then 4 from A' OR 0 from A and 5 from A'

    P(x<2)=(0.3)(0.7)^4+(0.7)^5=0.2401

    (2) I did not consider the one member who has no union because it doesn't make a difference so Union A - 4 , B - 0 , OR A-3 , B-1

    P(A>B)=(0.3)^4+(0.3)^3(0.6)=0.0243

    But unfortunately both are wrong . The answer given is (1) 0.528 (2) 1/9

    (a) Denote the X_i to equal 1 if the ith pick is from member A, and 0 otherwise. then we're interested in P(\sum_{i=0}^5{X_i}\leq1). If you then note that the X_i are (5,.3) binomial random variables, you compute
    P(\sum_{i=0}^5{X_i}\leq1) = \sum_{k=0}^{1}\binom{5}{k}(.3)^k(.7)^{5-k} = .52822

    (b)We're now interested in:
    P(\sum_{i=0}^4{X_i}=3|I) +  P(\sum_{i=0}^4{X_i}=4|I)

    where I is the condition that all samples consist of union workers from A or B.
    Again this can be solved using the binomial probability distribution, but with an adjusted probability of success equal to: 3/10*10/9=1/3. Using this, we get

    \Bigg{(}P(\sum_{i=0}^4{X_i}=3|I) +  P(\sum_{i=0}^4{X_i}=4|I)\Bigg{)} = \binom{4}{3}(1/3)^3(2/3)+\binom{4}{4}(1/3)^4 = \frac1{9}

    P.S. by the way, P(A|I) = \frac{P(A\cap{I})}{P(I)} = \frac{P(A\cap({A\cup{B}}))}{1-P(I^c)} = \frac3{10}*\frac{10}{9}
    Last edited by vince; February 21st 2010 at 09:01 AM. Reason: added the P.S.
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  3. #3
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    Hello, thereddevils!

    Here's the first one.
    (I'm still working on the second one.)


    In a large factory with many workers, 30% of the workers belong to union A,
    60% to Union B, and the remaining 10% didn't join any union.
    A sample of 5 workers is selected.

    (1) Find the probability that the sample has less than 2 members of Union A.
    "Less than two A's" means "no A's or one A."

    . . P(\text{no A's}) \;=\;9.07)^5 \:=\:0.16807

    . . P(\text{one A}) \:=\:{5\choose1}(0.3)^1(0.7)^4 \:=\:0.36015


    Therefore: . P(\text{less than two A's}) \;=\;0.16807 + 0.36015 \;=\;0.52822

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