Question 306204: 2. A bag contains two red balls, three blue balls and five green balls. Three balls are drawn at random. Find the probability that
a) the three balls are of different colours.
b) two balls are of the same colour.
c) all the three are of the same colour. 10 marks
Answer by toidayma(44)   (Show Source):
You can put this solution on YOUR website! Let nCk = number of ways to pick up k items from a set of n items.
Of course you should already know that  (*)
Bag consists of 2 red balls (R), 3 blue balls (B) and 5 green balls (G)
How many ways to pick up any 3 balls from the bag: 10C3
a> three drawn balls are of different colors --> this means to pick 1R, 1B and 1G
--> there are totally of 2C1*3C1*5C1 --> probability of 2C1*3C1*5C1/10C3
b> two of three drawn balls are of the same color.
There are 3 cases: the two balls of same color are red(1), blue (2) and green(3).
(1) means 2R and any one out of the 8 other balls (3B and 5G) --> 2C2*8C1
(2) means 2B and any one out of the 7 other balls (2R and 5G) --> 3C2*7C1
(3) means 2G and any one out of the 5 other balls (2R and 3B) --> 5C2*5C1
Thus, there are totally 2C2*8C1 + 3C2*7C1 + 5C2*5C1 ways to pick out 3 balls, 2 of which are of the same color --> the probability of (2C2*8C1 + 3C2*7C1 + 5C2*5C1)/10C3
c> all the three balls are of the same color.
This one is a bit more interesting. Obviously, the three drawn balls can't be red since there are only 2 red balls. Thus, there are only 2 cases: all three balls of the same color are blue(4) and green(5)
(4) means to pick 3B out of the 3 blue balls --> 3C3
(5) means to pick 3G out of the 5 green balls --> 5C3
Thus, there are totally 3C3 + 5C3 ways to pick out 3 balls, 2 of which are of the same color --> the probability of (3C3 + 5C3)/10C3
You surely can use the (*) for the final calculations, right?
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