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 $= 5×14×14+14×5×14+14×14×5=2940$
Probability $=\dfrac{2940}{6859}$
Is my understanding right? please correct if I am wrong.
Your answer works. Another way to solve this is to note that because draws are done with replacement, the draws are independent and identically distributed. Thus the random variable $X=\text{number of red balls chosen in 3 draws}$ is a binomial random variable with $n=3$ and $p=\text{probability of red ball on a given draw}=\frac{5}{5+6+8}=\frac{5}{19}$. Thus
$$P(X=1)=\binom{3}{1}\cdot \left(\frac{5}{19}\right)^1\cdot \left(1-\frac{5}{19}\right)^2= \frac{2940}{6859}.$$
Total number of ways $=19^3=6859$
Favorable ways $= 5×14×14+14×5×14+14×14×5=2940$
Probability $=\dfrac{2940}{6859}$
