A bag contains 3 red, 7 green and 4 yellow balls. 4 balls are drawn at random, find the probability that out of four balls at least 3 balls are yellow.

\(\frac{41}{1001}\)

Red = 3

Green = 7

Yellow = 4

Total =14

Reqd. probability = \(\frac{(^{4} \mathrm{ C }_3 \;\times \;^{10} \mathrm{ C }_1) \;+ \;^{4} \mathrm{ C }_4}{^{14} \mathrm{ C }_4}\)

                           = \(\frac{4\;×\;10\;+\;1}{1001}\) = \(\frac{41}{1001}\)

 Hence, option (C) is correct.