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A box contains 2 blue, 2 yellow, 2 green, 2 white, 2 purple and 2 red balls.

If 4 balls are randomly selected without replacement, what is the probability that all the selected balls are of different colours?

Any hint as to how to do this question?

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  • $\begingroup$ Are the balls of a given colour identical? $\endgroup$ – GoodDeeds Mar 6 '16 at 12:59
  • $\begingroup$ yes i think we can assume that $\endgroup$ – hazard Mar 6 '16 at 13:00
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Let each pair of balls of a particular colour be in a box of that colour.

Since the balls of a given colour are identical, choosing $4$ balls of different colours is equivalent to choosing four out of the given six boxes. This can be done in $6\choose4$ ways.

Among the unfavourable outcomes, it is possible that exactly two balls are of the same colour. Then, only three boxes would be selected and the pair of balls would be taken out from one of them. This can be done in $6\choose3$ $\times$ $3\choose1$ ways.

Finally, it is possible to get two pairs of balls of the same colour. This can be done by selecting any two boxes and picking out the pairs from both of them. This can be done in $6\choose2$ ways.

Thus, the probability of getting four balls of distinct colours is $$\text{Probability}=\frac{\binom{6}{4}}{\binom{6}{4}+\binom{6}{3}\times\binom{3}{1}+\binom{6}{2}}=\frac{15}{15+20\times3+15}=\frac{15}{90}=\frac16$$


Edit:

It turns out that the balls of a given colour are not supposed to be identical.

Then, the number of ways of choosing any four balls from the twelve distinct balls is $12\choose4$.

For the balls to have distinct colours, the number of ways to choose four boxes as above is $6\choose4$. From each box, we have a choice of two balls. Thus, the total number of ways of picking four balls with different colours is $\binom{6}{4}\times2^4$.

Then, $$\text{Probability}=\frac{\binom{6}{4}\times2^4}{\binom{12}{4}}=\frac{48}{99}=0.4848...$$

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  • $\begingroup$ Thanks for the answer! but the actual answer given is 0.4848 $\endgroup$ – hazard Mar 6 '16 at 13:13
  • $\begingroup$ @hazard That's the answer if the balls are not identical $\endgroup$ – GoodDeeds Mar 6 '16 at 13:18
  • $\begingroup$ @hazard Added that to the answer. $\endgroup$ – GoodDeeds Mar 6 '16 at 13:23
  • $\begingroup$ You inverted the fraction $$\frac{\binom{6}{4} \cdot 2^4}{\binom{12}{4}}$$ in the second interpretation of the problem. $\endgroup$ – N. F. Taussig Mar 6 '16 at 13:35
  • $\begingroup$ @N.F.Taussig Thank you for the correction $\endgroup$ – GoodDeeds Mar 6 '16 at 13:42
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Just an alternative way of thinking:

First it doesn't matter which ball you pick. Then after that you have $\frac{10}{11}$ ways to pick the next ball. For the third ball you have $\frac{8}{10}$ ways to pick it. And for the last ball you have $\frac{6}{9}$ ways to pick it. Multiplying these gives you: $$ 1 \cdot \frac{10}{11} \cdot \frac{8}{10} \cdot \frac{6}{9} = \frac{16}{33} = 0.4848... $$

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