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Probability of stealing a base Question

  1. May 7, 2008 #1
    I have been grappling with this one for the past 24 hours.

    I'll make this baseball related, consisting of three independent events assuming the following:
    Player A has a 40% chance of stealing a base in "his game" tonight.
    Player B has a 60% chance of stealing a base in "his game" tonight.
    Player C has a 10% chance of stealing a base in "his game" tonight.

    I want to figure out what the chances are that AT LEAST ONE of the three players will steal a base tonight. I originally thought the formula was P(AUBUC)=P(A)+P(B)+P(C)-P(A)*P(B)*P(C), but that gives me a final result of 1.076... a result higher than 100%. So that can't be it.

    Can somebody help me out with this one?
  2. jcsd
  3. May 7, 2008 #2


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    The formula for three items should be written as:
    P(A[tex]\cup[/tex]B[tex]\cup[/tex]C)=P(A) + P(B) + P(C) - P(A) P(B) - P(A) P(C) - P(B) P(C) + P(A) P(B) P(C) = 0.784
  4. May 7, 2008 #3
    Another way to do it is this:

    Let S represent the number of steals.

    P(S>=1) =
    1 - P(S<1) =
    1 - P(S=0) =
    1 - P(not A and not B and not C) =
    1 - P(not A)*P(not B)*P(not C) <---- Because they are independent events.

    Now, P(not A) = 1-.4 = .6, P(not B) = .4, P(not C) = .9

    So, P(S>=1) = 0.784
  5. May 7, 2008 #4
    Thank you both for your responses and thank you for helping me find out how to solve such problems. I like this second method since it is easier to expand to more events by simply adding more "P(not D), P(not E), etc."

    Thanks again.

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