A bag contains 4 red, 5 white and 6 black balls. Three balls are selected from this bag simultaneously. The probability that one of the colour will be missing in the selected balls, is equal to

$\frac{301}{455}$

(i) Selected balls can be W W B or W B B (when red colour is missing).

Corresponding probability

$=\frac{{ }^5 C_2 .{ }^6 C_1+{ }^5 C_1 .{ }^6 C_2}{{ }^{15} C_3}=\frac{135}{{ }^{15} C_3}$

(ii) Selected balls can be R R W or R W W (when black colour is missing).

Corresponding probability

$=\frac{{ }^4 C_2 .{ }^5 C_1+{ }^4 C_1 .{ }^5 C_2}{{ }^{15} C_3}=\frac{70}{{ }^{15} C_3}$

(iii) Selected balls can be RBB, RRB (when white colour is missing).

Corresponding probability

$=\frac{{ }^4 C_1 .{ }^6 C_2+{ }^4 C_2 .{ }^6 C_1}{{ }^{15} C_3}=\frac{96}{{ }^{15} C_3}$

Thus, required probability = $\frac{301}{455}$