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.
Answer & explanation
Correct answer: option 3
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.