# Oliver has 30 marbles, 12 are red, 10 are green and 8 are black. he asks three of his friends to take out a marble and replace it. what is the probability that his friends each take out a different colored marble?

Mar 24, 2017

To be checked

#### Explanation:

Let the probability of a color be designated as $P \left(\text{color}\right)$

Let red be R $\to P \left(R\right) = \frac{12}{30}$

Let green be G $\to P \left(G\right) = \frac{10}{30}$
Let black be B $\to P \left(B\right) = \frac{8}{30}$

These probabilities do not change as you progress through the selection as what is selected is returned to the sample.

$\cancel{\text{Each person selects 3 and returns after each selection. }}$

Each person selects 1 and returns it ready for the next person to make their selection.

$\textcolor{b r o w n}{\text{All possible success type selection:}}$

Note that this diagram is only for the 'success' part. To include the fail part would make the diagram rather large.

So the probability is:

$6 \times \left[\frac{8}{30} \times \frac{10}{30} \times \frac{12}{30}\right] = \frac{16}{75}$

Mar 24, 2017

16/75 or 21.3%

#### Explanation:

We can break this down into two steps. First, what is the probability of three different coloured balls being chosen?

Since the ball is being replaced each time, this is simple. The chances of choosing a red ball are 12/30, those of choosing a blue ball are 10/30 and those of choosing a black ball 8/30. The probability therefore of choosing three different coloured balls is the product of each probability, the order is immaterial. This is therefore (12/30)x(10/30)x(8/30).

Now, we have to work out how many ways there are of choosing three different coloured balls. This comes out at 3 factorial i.e. 3x2x1 = 6. This is because there are three ways of choosing the first ball i.e. red or green or black, but only two ways of choosing the second (because we have already chosen one colour so there are only two colours left, as each ball has to be a different colour) and only one way of choosing the last (by the same argument).

The overall probability therefore is 6 times the probability of choosing three different coloured balls (6x(12/30)x(10/30)x(8/30)), which comes out to the number given above.