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An urn contains seven red balls, seven white balls, and seven blue balls. A sample of five balls is drawn at random without replacement. Compute the probability that:

E = The sample contains at least one ball of each colour.

The answer is:

$$P(E) = \frac{C(3,1)×C(7,3)×C(7,1)×C(7,1)+C(3,2)×C(7,2)×C(7,2)×C(7,1)}{C(21,5)}$$

But I don't get how they calculated the numerator, what are the two separate events?

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The events are:

  • three balls of one colour, and one each of the other two.
  • two balls of two colours, and one of the remainder.

We count the ways to select which colours belong to the group, then select balls for each.

  • $\binom 3 1$ ways to select a colour, $\binom 73$ ways to select three balls of that colour, then $\binom 7 1\binom 71$ ways to select one ball of each of the other colours.
  • $\binom 3 2$ ways to select two colours, $\binom 72\binom 72$ ways to select two balls of each of those colours, then $\binom 7 1$ ways to select one ball of each of the remaining colour.

$$\dfrac{\binom 3 1\binom 73\binom 7 1\binom 71+\binom 3 2\binom 72\binom 72\binom 7 1}{\binom{21}{5}}$$

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