# An urn contains nine balls, two of which are red, three blue and four black. Three balls are drawn from the urn at random. What is the probability that

99 views (i) the three balls are of different colours?

(ii) the three balls are of the same colour?

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we have Red = 2   Blue =3   Black = 4         Total = 9 balls. Therefore n= 9

3 balls will be drawn out of the 9.  we can represent this as : 9C

Hence finding the sample space n(s) we have   9C3 =  9! / (9-3)! * 3!  Which is equal to 84

Now to solve the questions.

(i) In order to solve for the first question we need the answer from the second question, so we solve the second one first.

(ii) The 3 balls are the same colorLet X be the event of getting same color.  This means all 3 are RED or all 3 are BLUE or all 3 are BLACK. and we know the numbers of each color which are 2 ,3 and 4 for red, blue and black respectively. Thus we can write the sample space of x as:  n(X) = 2C3 + 3C3 + 4C

Note that  2C3  = 0. (Why? we want to find the probability of picking 3 balls that are the same color, but for the color RED there are only 2 balls that are RED  out of the 9 total balls so it would be impossible to pick 3 RED when there are only 2 RED so answer is 0).

3C3 = 1  And   4C3 = 4 ( formula used n! /(n-r)! * r! ).

n(X) : 2C3 + 3C3 + 4C3   = 0 + 1 + 4 = 5.

From the Probability formula we then have that  p(X) = n(X) n(S) = 5/84

Now back to question number (i) to find the answer all we need to do is subtract 1 from the probability of getting all 3 balls the same color, and we already found that in (ii).

Therefore we have  1- 5/84  =  79/84

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