A car repair garage has two spare sets of spark plugs in stock for a particular model of car. Each time one of these cars comes in for repair, there is a probability of $0.05$ (independently on each occasion) that the spark plugs will need to be replaced so that one of the spare sets will be used.
(a) Let $X$ denote the number of these cars that can be repaired before the two sets of spark plugs have been used (including the car to which the final set is fitted). Write down the mean and variance of $X$.
(b) What is the probability that 10 or more cars can be repaired before the final set of spark plugs is used?
For a) am I right thinking that $X$ is distributed binomially with parameters $n$ and $0.05^2$? If So, I am not sure how I can calculate $P(X\geq 10)$ for part (b)
edit:
a): $X$ is distributed negative binomial with parameters $2$ and $0.05$. E(X) = 50 and var(X) = 760.
b): I am not sure how to do this other than summing $P(X\geq 10) = 1- P(X \leq 9)$
a) Not quite, the expectation is $$E[X] = E[X_1+X_2] = 2E[X_1] = \frac{2}{.05} = 40,$$ where $X_i\sim\text{Geom}(p = .05)$ on $\{1,2,3\dotsc\}$, and I use that fact that the sum of (two) independent geometrics is a negative binomial.
b) Notice that, regarding this car model, $$A = \{\text{You repair 10 or more cars}\}\iff B=\{\text{You only use 0 or 1 sets in 9 car repairs}\}$$ Thus, it is easier to do $$P(X\geq 10) = P(A) = P(B).$$ So \begin{align*} P(B) &= P(\text{Use 0 sets in 9 repairs})+P(\text{Use 1 set in 9 repairs})\\ &= \binom{9}{0}.05^0(1-.05)^9+\binom{9}{1}.05^1(1-.05)^8\\ &=0.9287886 \end{align*}