An urn contains $5$ red, $6$ blue and $8$ green balls. $3$ balls are randomly selected from the urn, find the probability of getting exactly one red ball if the balls are drawn with replacement.
Source: doubt came from the similar question here.
In the same line, my answer follows
Total number of ways $=19^3=6859$
Favorable ways $= 5×14×14+14×5×14+14×14×5=2940$
Is my understanding right? please correct if I am wrong.