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B OR probabilities

  1. Apr 28, 2016 #1
    A code is generated consisting of 3 letters followed by 2 digits. Each of the 3 letters generated is equally likely to be any of the 26 letters of the alphabet. Each of the 2 digits generated is equally likely to be any of the nine digits 1-9. Find the probability that a randomly chosen code has exactly 2 letters the same or 2 digits the same, but not both.

    Attempted solution:

    P(exactly 2 digits same) = (263*9)/(263*92) = 1/9
    P(exactly 2 letters same) = (26*25*3*92)/(263*92) = 75/676
    P(2 digits same AND 2 letters same) = (26*25*3*9)/(263*92) = 25/2028
    Required probability = 1/9 + 75/676 - 25/2028 = 0.210

    But the correct answer is 0.197. Can I ask where did I go wrong? I can't figure it out...
    Thanks!
     
  2. jcsd
  3. Apr 28, 2016 #2

    stevendaryl

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    Your calculation for 2 letters is the probability that at LEAST 2 letters is the same. You have to subtract off the probability that all 3 letters are the same to get the probability that exactly 2 letters are the same.
     
  4. Apr 28, 2016 #3
    One other possible mistake. Exactly how many ways are there to have 2 items the same in a set of 3?
     
  5. Apr 28, 2016 #4
    But why is that so?
    __ __ __ __ __
    For two letters to be the same == 26 choices
    For the last letter, since it can't be the same as the two letters that are the same == 26 - 1 = 25 choices
    No. of ways to arrange 3 letters, when 2 are the same == 3!/2! = 3
    Thus no. of ways == 26 * 25 * 3 * 92

    Why did I calculate the probability that at least 2 letters are the same? Could you elaborate? Sorry kind of confused.
     
  6. Apr 28, 2016 #5
    Yup I'm dumb sorry :D
     
  7. Apr 28, 2016 #6

    stevendaryl

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    Hmm. My criticism was mistaken, but here's a suggestion: Instead of computing probabilities for the various possibilities, try just computing

    [itex]N_L =[/itex] the number of ways to have exactly 2 letters the same, and both digits different.
    [itex]N_D =[/itex] the number of ways to have 3 or 0 letters the same, and both digits the same.

    Then the probability is [itex]\frac{N_L + N_D}{26^3 \cdot 9^3}[/itex]

    When I did it that way, I got the expected answer.
     
  8. Apr 28, 2016 #7

    stevendaryl

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    Okay, I figured out what you did wrong:

    You used [itex]P(A \vee B) = P(A) + P(B) - P(A \wedge B)[/itex] where [itex]A[/itex] = "exactly 2 letters the same" and [itex]B[/itex] = "both digits the same". But they aren't asking for [itex]A \vee B[/itex], they're asking for [itex]A\ xor\ B[/itex] ([itex]xor[/itex] = exclusive or). The formula for that is:

    [itex]P(A\ xor\ B) = P(A) + P(B) - 2P(A \wedge B)[/itex]
     
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