Detailed Answer
(C)
Out of 9, 3 (1 black & 2 red. are expected to be drawn)
Hence sample space
n(S) = 9c3
= 9!/(6!×3!)
= 362880/4320
= 84
Now out of 4 black ball 1 is expected to be drawn hence
nb. = 4c1
= 4
Same way out of 5 red balls 2 are expected be drawn hence
n(R) = 5c2
= 5!/(3!×2!)
= 120/12
= 10
Then P(B U R) = n(B)×n(R)/n(S)
i.e 4×10/84 = 10/21