In a basket of fruits, there are 6 grapes, 11 bananas and 13 oranges, if one fruit is chosen at random, what is the probability that the fruit is either a grape or a banana?
\(\frac{17}{30}\)
\(\frac{11}{30}\)
\(\frac{6}{30}\)
\(\frac{5}{30}\)
Correct answer is A
Pgrape or Pbanana = \(\frac{6}{30}\) + \(\frac{11}{30}\)
= \(\frac{17}{30}\)