We know that diskettes made by a factory have defects with probability $0.3$, independently of each other. The factory sells packages containing $10$ diskettes and offers a guarantee that at most one of the $10$ is defective and states that it will refund the purchase money if the number of defective products exceeds this warranty number. If we buy $3$ packages, what is the probability that one of them will be returned?
MY ATTEMPT
Let us consider a package. If we denote by $X$ the random variable which counts the number of defective diskettes in a package, we conclude that $X\sim B(10,0.3)$. According to this notation, the probability that at most one diskette is defective in the package is given by \begin{align*} \textbf{P}(X\leq 1) = \textbf{P}(X = 0) + \textbf{P}(X = 1) = {10\choose 0}\left(\frac{3}{10}\right)^{0}\left(\frac{7}{10}\right)^{10} + {10\choose 1}\left(\frac{3}{10}\right)^{1}\left(\frac{7}{10}\right)^{9} \end{align*}
For convenience, we set $p = \textbf{P}(X\leq 1)$. Therefore the probability that exactly one of the three packages will be returned is given by $3p^{2}(1-p)$.
Am I on the right track? I would like to know somebody else's opinion on the subject. Thanks!