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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
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The correct answer is:
$\frac{261}{286}$
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]
$=1-\frac{25}{286}=\frac{261}{286}$
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