Probability problem on punctured wheels 
Someone drives a car with four new wheels. In the trunk of the car there is a spare wheel. The probability that a new wheel will have a puncture during the drive is $0.05$. The probability that the spare wheel will have a puncture during the drive is $0.25$.
During the course of the drive, one of the (new) wheels had a puncture, and the driver replaced it with the spare wheel. What is the probability that, after the first punctured wheel was replaced, exactly one other wheel will be punctured during the drive?

Since the type of wheel was not specified, I calculated the probability of a wheel being punctured after the spare wheel was installed as the sum of probabilities that a new wheel and the spare wheel will be punctured: $0.05+0.25=0.3$. After that, I used the binomial distribution formula:
$$ \binom{4}{1}\cdot 0.3^1\cdot 0.7^3 = 0.4116 $$
However, according to the textbook, the answer is $$ \frac{2527}{8000} = 0.315875 $$
What am I doing wrong? And how can this problem be solved? Help will be much appreciated.
 A: There are in fact two possibilities to consider, given that there are now three new wheels and one spare wheel, for exactly one wheel to puncture:


*

*The spare wheel punctures and the other three wheels don't. This has a probability $\frac14×\left(\frac{19}{20}\right)^3=\frac{6859}{32000}$.

*One of the three new wheels punctures while the spare wheel and the other two new wheels don't. This has a probability of $3×\frac34×\frac1{20}×\left(\frac{19}{20}\right)^2=\frac{3249}{32000}$.


Here $\frac34$ and $\frac{19}{20}$ are complements, probabilities that a spare and new wheel will not puncture respectively.
Adding the probabilities of the two cases together, we find a $\frac{6859+3249}{32000}=\frac{2527}{8000}$ chance of exactly one wheel puncturing, as given.
A: $$\begin{align}\text{Probability}&=\text{P}(\text{spare wheel punctures})+3\times\text{P}(\text{new wheel punctures})\\&=0.25\times\underbrace{(1-0.05)^3}_{\text{new wheels}\\\text{didn't puncture}}+3\times\underbrace{(1-0.25)}_{\text{spare wheel}\\\text{didn't puncture}}\times0.05\times\underbrace{(1-0.05)^2}_{\text{other two wheels}\\\text{didn't puncture}}\\&=0.315875\end{align}$$
