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    Working with conditional probability.

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    There are 12 light bulbs, 5 of which are defective. All bulbs look alike and have an equal probability of being chosen. Three light bulbs are selected and placed on the table.

    Find the probability that:

    a) All three are defective.

    b) Two are defective.

    c) That none are defective.

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    Solution Preview

    Assuming no replacement
    Let defective bulbs be D
    non defective bulbs be ND

    a) P(all 3 defective) = D x D x D
    = ...

    Solution Summary

    The probability of picking defective light bulbs is calculated.