A bag contains 10 balls of which 2 are red and the remaining are either blue or black. If the probability drawing 3 balls of the same colour is 1120 and if the  number of blue balls exceeds the number of black balls, the number of blue balls is

Let the number of blue balls be n. ∴  the number of black balls =10−(n+2)=8−n.  If all the three balls drawn are of the same colour, they must  be either blue or black,  the probability for which is  nC3 10C3+ 8−nC3 10C3=11120⇒ n(n−1)(n−2)+(8−n)(7−n)(6−n) =10×9×8⋅11120⇒ 18n2−144n+336=66⇒ n2−8n+15=0⇒ (n−5)(n−3)=0⇒ n=5 (as n> number of black balls)