A box contains 6 bottles of $\mathrm{V}_1$ drink, 3 bottles of $\mathrm{V}_2$ drink and 4 bottles of $\mathrm{V}_3$ drink. If three bottles are drawn at random, then the probability that the three are not of the same variety is

• A

  $\frac{632}{713}$
• B

  $\frac{752}{833}$
• C

  $\frac{833}{858}$
• Correct answer
  D

  $\frac{261}{286}$

Solution

P[All of three bottles are of same variety] $=\frac{6_{\mathrm{C}_3}+3_{\mathrm{C}_3}+4_{\mathrm{C}_3}}{13_{\mathrm{C}_3}}=\frac{\frac{6.5 \cdot 4}{3.2 .1}+1+4}{\frac{13 \cdot 10.11}{3.2 .1}}$ $P(E)=\frac{25}{26 \times 11}$ $\therefore \mathrm{P}[$ All of three bottles are not of same variety] $=1-\frac{25}{286}=\frac{261}{286}$

Hence, the correct answer is option: 'D'.