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statisticsstatsd

lever 5 times that you will get your 3rd Green Light on the 5th pull of the lever

Assume P(Red Light) = .40 and P(Green Light) = .60.

Solution

  • This is not answerable as stated; at least, not without assuming independence and without a clearer statement of what the probability is out of (the reference class: is it out of all ways in which the five pulls could turn out, or only the ones in which there are three green?)

    If we're willing to assume indendence and that we're talking about any combination of 5 pulls, then the desired outcome is the probability that:

    1. There are 3 green lights of 5 AND
    2. The fifth is green.

    This is the same as

    1. There are 2 green lights in the first 4 AND
    2. The fifth is green.

    The probability of getting 2 green lights of 4 can be calculated using the binomial probability distribution and is

    choose(4,2) * .4^2 * .6^2

    The probability of the fifth being green is .6. So the whole probability is

    choose(4,2) * .4^2 * .6^2 * .6 = choose(4,2) * .4^2 * .6^3