A box contains 6 bottles of variety 1 drink, 3 bottles of variety 2 drink and 4 bottles of variety 3 drink. Three bottles of them are drawn at random, what is the probability that the three are not of the same variety.

Answer: Option C

Total number of drink bottles = 6 + 3 + 4 = 13
Let S be the sample space
Then, n(S) = number of ways of taking 3 drink bottles out of 13
Therefore,
$$\begin{array}{rl}& \text{n}\left(\text{S}\right)={}^{13}{C}_3\\ & =\frac{13\times 12\times 11}{1\times 2\times 3}\\ & =22\times 13\\ & =286\end{array}$$
Let E be the event of taking 3 bottles of the same variety.
Then, E = event of taking (3 bottles out of 6) or (3 bottles out of 3) or (3 bottles out of 4)
$$\begin{array}{rl}& \text{n}\left(\text{E}\right)={}^6{C}_3+{}^3{C}_3+{}^4{C}_3\\ & =\frac{6\times 5\times 4}{1\times 2\times 3}+1+\frac{4\times 3\times 2}{1\times 2\times 3}\\ & =20+1+4\\ & =25\end{array}$$
The probability of taking 3 bottles of the same variety $$=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{25}{858}$$
Then, the probability of taking 3 bottles are not of the same variety
$$\begin{array}{rl}& =1-\frac{25}{286}\\ & =\frac{261}{286}\end{array}$$