Mathematics - Probability Question with Solution | TestHub

Since $a+b+c=6$, the possible digit selections are $(1,2,3),(1,1,4),(2,2,2),(0,1,5),(0,2,4),(0,3,3),(0,0,6)$

The required number of ways $6+3+1+4+4+2+1=21$

Required probability $=\frac{21}{9\times 10\times 10}=\frac{7}{300}$