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.

Solution(By Examveda Team)

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,
$$\eqalign{ & {\text{n}}\left( {\text{S}} \right) = {}^{13}{C_3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{13 \times 12 \times 11}}{{1 \times 2 \times 3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 22 \times 13 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 286 \cr} $$
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)
$$\eqalign{ & {\text{n}}\left( {\text{E}} \right) = {}^6{C_3} + {}^3{C_3} + {}^4{C_3} \cr & = \frac{{6 \times 5 \times 4}}{{1 \times 2 \times 3}} + 1 + \frac{{4 \times 3 \times 2}}{{1 \times 2 \times 3}} \cr & = 20 + 1 + 4 \cr & = 25 \cr} $$
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
$$\eqalign{ & = 1 - \frac{{25}}{{286}} \cr & = \frac{{261}}{{286}} \cr} $$