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Probablity :confused totally

  1. Oct 23, 2006 #1


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    Hi,Can anyone help me with this one?

    Probability of an event occuring is 0.6

    1)Probability of 1 of such events occurring out of total 5?

    2)and 4 of such event occurring out of total 5.?

    Answers given are:

    1) 0.768
    2) 0.2592

    Please help me out by giving and explaining this one?..
    thnks in advance
  2. jcsd
  3. Oct 23, 2006 #2


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    Homework Helper

    You may want to use the binomial distribution [tex]f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k}[/tex], where n is the number of trials, and k the number of times an event whose probability is 0.6 occured.
  4. Oct 23, 2006 #3


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    I tried..but still not getting the ans..dunno where am going wrong..could u pls solve it and show me...
    thnk u very much!....i guess i am making the same mistake again n again..but cant see thro ' it..will really appreciate if u solve it and explain...

    U r right tht we have to use binomial distribution..is there any other way also to solve it?

  5. Oct 23, 2006 #4


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    Complete words, please!

    I'm not sure at all where you're going wrong. You have the formula -- just plug in the appropriate values and you have the answer. What are you getting, and how?
  6. Oct 23, 2006 #5


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    You have already been told to use
    [tex]f(k)=\left( \begin{array}{c} n \\ k \end{array} \right) p^k (1-p)^{n-k}[/tex]

    When p= 0.6, n= 5, P(1)= [itex]\frac{5!}{(4!)(1!)}.6^1 .4^4[/itex]
    [itex]= 5(.6)(0.025)= 0.0768[/itex]. Was it the arithmetic you had trouble with?

    To answer (2) take k= 4 rather than 1.
  7. Oct 23, 2006 #6


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    sorry frnds..was making a very silly mistake with decimals..I got it after radou's 1st reply..was just trying to work out if there is any other way to solve besides using the binomial formula..
    anyways, i have stuck with wht u all suggest..thnks again all of you!!..appreciate it!:smile:
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