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I am fairly new to statistics and I would appreciate any help on this problem.

The length of a large batch of 1-inch nails has a normal distribution with μ = 1 and σ = 0.08

If a nail has a length < 0.9 it's too short, and if it has a length of > 1.15 it's too long.

If we pick 10 nails randomly, what's the probability that exactly one nail is too short and one nail is too long?

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    You have a multinomial distribution with $n=10$, $p_1=\Phi(-0.1/0.08)$, $p_3=1-\Phi(0.15/0.08)$, $p_2=1-p_1-p_3$ where $\Phi$ is the standard normal CDF. You're then asking for the probability that $n_1=1,n_3=1,n_2=8$. – 2017-01-12

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There are three categories of nails, too short, too long, and acceptable. The probabilities of drawing each are given by \begin{align*} p_s &= \Phi_{\mu,\sigma^2}(.9) \\ p_l &= 1 - \Phi_{\mu,\sigma^2}(1.15) \\ p_a &= \Phi_{\mu,\sigma^2}(1.15) - \Phi_{\mu,\sigma^2}(.9) \end{align*} Converting to the CDF of a standard normal, we have \begin{align*} \frac{X - \mu}{\sigma} \stackrel{(d)}{=} N(0,1) \Rightarrow \frac{X - 1}{.08} \stackrel{(d)}{=} N(0,1) \end{align*} So our probabilities can be written \begin{align*} p_s &= \Phi(-.1 / .08) \approx .106 \\ p_l &= 1 - \Phi(.15/.08) \approx .03 \\ p_a &= \Phi(.15/.08) - \Phi(-.1/.08) \approx .864 \end{align*} Thus the probability that we get exactly 1 short nail, 1 long, and the rest acceptable is \begin{align*} Pr = \frac{10!}{1! 1! 8!} p_s p_l p_a^{8} \approx 90 \cdot (0.106) \cdot (0.03) \cdot (0.864)^8 = 0.0897 \end{align*}