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    hunt68's Avatar
    hunt68 Posts: 1, Reputation: 1
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    #1

    Jun 28, 2012, 05:35 AM
    Probability of 3 out of 4 independent events occurring
    I have 4 golfers playing together in a foursome. I know the probability of each making a par or better on a given hole:
    A-player: 60%
    B-player: 50%
    C-player: 40%
    D-player: 30%
    What is the probability that any three of these 4 golfers will make par or better on the hole?

    ANSWER: Need help on how to calculate this. Is it just adding together the four probabilities of the possible outcomes where 3 players shoot par or better? Namely: P(A&B&C)+P(A&B&D)+P(A&C&D)+P(B&C&D) = .12+.09+.072+.06 = 34.2%. Is this it or am I missing something?
    ebaines's Avatar
    ebaines Posts: 12,132, Reputation: 1307
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    Jun 28, 2012, 05:55 AM
    Quote Originally Posted by hunt68 View Post
    Is it just adding together the four probabilities of the possible outcomes where 3 players shoot par or better? Namely: P(A&B&C)+P(A&B&D)+P(A&C&D)+P(B&C&D) = .12+.09+.072+.06 = 34.2%. Is this it or am I missing something?
    Remember that for exactly 3 players to make par the 4th player must not make par. For example the probability that A, B and C make par and D does not is P(A&B&C&not D) = (.6)*(.5)*(.4)*(1-.3) = 0.084. By leaving out the (1-.3) term what you have calculated is the probability that players A, B, and C make par without caring whether D makes par or not, so it includes the possibility that all 4 players make par. Hope this helps - try it again and post back with your answer.

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