1. ## Probability on balls

Q1: A bag contains 5 red balls, 5 green balls, and 6 blue balls. If 3 balls are removed at random and no ball is returned to the bag after removal, what is the probability that all 3 balls are red? Answer should be in common fraction.

To this problem, I tried....
5/16 * 4/15 * 3/14 = 1/56
Is my approach correct.

Also, I am confused to solve the above problem in the following method?
16c3 * 15c2 * 14c1

Also wants to know, which practice is best to follow?

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Q2: A bag contains 3 red balls, 3 green balls, and 4 blue balls. If 2 balls are removed at random and each ball is returned to the bag after removal, what is the probability that all 2 balls will be red?

3/10 * 3/10 = 9/100
or
10c3 * 10c3
I am not sure about the second method?

Thanks,

2. Originally Posted by Ka9
Q1: A bag contains 5 red balls, 5 green balls, and 6 blue balls. If 3 balls are removed at random and no ball is returned to the bag after removal, what is the probability that all 3 balls are red? Answer should be in common fraction.

To this problem, I tried....
5/16 * 4/15 * 3/14 = 1/56
Is my approach correct. Mr F says: Yes.

Also, I am confused to solve the above problem in the following method?
16c3 * 15c2 * 14c1 Mr F says: Why try doing it in a way that requires much more computation and time.

Also wants to know, which practice is best to follow? Mr F says: The way that you feel most comfortable with in getting the correct answer.

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Q2: A bag contains 3 red balls, 3 green balls, and 4 blue balls. If 2 balls are removed at random and each ball is returned to the bag after removal, what is the probability that all 2 balls will be red?

3/10 * 3/10 = 9/100 Mr F says: Correct.
or
10c3 * 10c3
I am not sure about the second method? Mr F says: Then do not complicate things by using this approach.