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Probability Question

  1. Feb 19, 2015 #1
    Its number 6 but im unsure of my answer...
    This was my attenpt at the question
     

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  2. jcsd
  3. Feb 20, 2015 #2
    yes i got the same answer.It is (4/10)*(8/11)+(6/10)*(4/11)=28/55.(The first term for getting a red from first bag and again a red from the second bag and second term for picking a black in first bag and again picking a black in second bag)
     
  4. Feb 21, 2015 #3
    Ahh thanks man
     
  5. Feb 22, 2015 #4
    Guys im stuck on the first part of part C on question 11. The next picture shows what I came up with but I have no idea how to continue it. The answer is 810 but all my attempts give me an answer over 1000 :\
     

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  6. Feb 26, 2015 #5

    HallsofIvy

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    11 (c) says "There is at least one repeated digit, but no digit appears more than twice in a number. Find the probability that a four figure number chose at random from the from the set of numbers in case (a) above contains at least one 6".

    I find that very confusing. The first sentence does not appear to have anything to do with the problem posed in the second sentence!

    Problem 11(a) you calculated how many such numbers there were. Now calculate how many of them do NOT have a "6". The answer to the question asked in 11(c) is 1 minus (the number that have no "6" divided by the answer to (6)).
     
  7. Feb 26, 2015 #6

    Ray Vickson

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    Please try to respect PF standards, by typing up the questions and your answers. Your solution image comes out sideways, and I am unable to rotate it.
     
  8. Feb 27, 2015 #7

    haruspex

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    I assume the last sentence in 11c is supposed to be a completely separate question.
    Your attached working is impossible to decipher. Please explain your approach to 11c (first part).
     
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