# 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.

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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