A bag contains 10 red balls, 10 green balls and 6 white balls. Two balls are drawn at random from th

Discussion in 'Calculator Requests' started by math_celebrity, Jan 8, 2017.

  1. math_celebrity

    math_celebrity Administrator Staff Member

    A bag contains 10 red balls, 10 green balls and 6 white balls. Two balls are drawn at random from the bag without replacement. What is the probability that they are of different colours?
    • The key phrase here is without replacement.
    • First, it's easier to find the probability of both colors matching, and then subtracting that from 1.
    We want 1 - (P(Red-Red) + P(Green-Green) + P(White-White)). So we have the following:

    Find the probability of both colors matching
    P(Red-Red) = 10/26 * 9/25 = 0.138462
    P(Green-Green) = 10/26 * 9/25 = 0.138462
    P(White-White) = 6/26 * 5/25 = 0.046154
    P(Red-Red) + P(Green-Green) + P(White-White) = 0.13846 + 0.13846 + 0.046154 = 0.3231

    Now, we want to take the complement of this probability which is no colors matching, so we have:
    P(Both Different Colors) = 1 - P(Same Colors)
    P(Both Different Colors) = 1 - 0.3231
    P(Both Different Colors) = 0.6769

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