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 : ^{9}C_{3 }

_{Hence finding the sample space n(s) we have }^{9}C_{3 = 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 color** : **Let **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) = ^{2}C_{3} + ^{3}C_{3} + ^{4}C_{3 }

Note that ^{2}C_{3 = 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).}

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

n(X) : ^{2}C_{3} + ^{3}C_{3} + ^{4}C_{3 = 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**