A bag contains 4 red, 5 black and 6 blue balls. The probability that two balls drawn at random simultaneously from the bag are a red and a black ball is $\mathrm{\alpha }$ then $21\mathrm{\alpha }$ =___________

Total balls in bag=15Out of two balls can be drawn in ${ }^{15}{\mathrm{C}}_2=105$ ways.Out of  4 red balls, one ball can be drawn in ${ }^4{\mathrm{C}}_1=4$ways.
And out of 5 black balls one ball can be drawn in ${ }^5{\mathrm{C}}_1=5$ ways
If $E$ is the event of getting a red and a black ball in one draw, the no of favourable to  $E=4\times 5=20\therefore$  Probability is $\frac{20}{105}=\frac{4}{21}=\mathrm{\alpha }\therefore 21\mathrm{\alpha }=4$