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AP EAMCET · Maths · Probability

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

  1. A \(\frac{632}{713}\)
  2. B \(\frac{752}{833}\)
  3. C \(\frac{833}{858}\)
  4. D \(\frac{261}{286}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{261}{286}\)

Step-by-step Solution

Detailed explanation

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…