1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help, i keep on getting probability >1

  1. Apr 9, 2005 #1
    Roll a fair die twice and find the probability of at least one 4
    here's what i did:

    P(A) = |A|/|Ω| = (6C1*6C0+6C1*6C1)/(6C1*6C1)
    And i get the answer as 42/36 and the real answer is 5/36
    what did I do wrong
     
  2. jcsd
  3. Apr 9, 2005 #2

    xanthym

    User Avatar
    Science Advisor

    Although enumeration isn't the most efficient solution to probability problems, sometimes it can show where errors are occuring. All possible 36 outcomes of 2 rolls of a fair die are shown below. Those containing At Least One (1) "Four (4)" are highlited. By counting the highlited cases below, it can be seen that exactly (11) cases out of (36) show at least one (1) "Four (4)". Therefore, the required probability is (11/36).

    Roll
    AB
    ==
    11
    12
    13
    14
    15
    16

    21
    22
    23
    24
    25
    26

    31
    32
    33
    34
    35
    36

    41
    42
    43
    44
    45
    46


    51
    52
    53
    54
    55
    56

    61
    62
    63
    64
    65
    66
    ==

    ~~
     
  4. Apr 9, 2005 #3
    huan,

    xanthym is right; it's hard to argue with a 100% coverage test!

    Another way to get the same answer (and it works on LOTS of probability problems) relies on the fact that "getting at least one 4" and "getting exactly zero 4s" are mutually exclusive and collectively exhaustive. In other words, all possible outcomes are included in one or the other but not both of these. So their probabilities have to sum to 1. But the probability of "getting exactly zero 4s" is easy to calculate; it's just 5/6 *5/6 = 25/36. So the prob of "getting at least one 4" is 1- 25/36 = 11/36
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?