A) $ \frac{41}{190} $
B) $ \frac{21}{190} $
C) $ \frac{59}{190} $
D) $ \frac{99}{190} $
E) $ \frac{77}{190} $
Correct Answer: D
$ \therefore $ $ n,(S)={}^{20}C _2=190 $ For same colour of ball, $ n,(E _1)={}^{13}C _2=78 $ and $ n,(E _2)={}^{7}C _2=21 $ Hence, required probability, $ P(E)=\frac{n,(E _1)}{n,(S)}+\frac{n,(E _2)}{n,(S)}=\frac{78}{190}+\frac{21}{190}=\frac{99}{190} $