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)


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 $P(X\geq 10) = 1- P(X \leq 9)$

  • $\begingroup$ I'm sorry, I think your answer of part a) is wrong. That's not binomial. I would suggest you to think it over. $\endgroup$
    – Max
    Commented Jan 27, 2016 at 22:53
  • 2
    $\begingroup$ No, this is a negative binomial distribution. If you are unfamiliar with that, then you can try to construct the probability from first principles. $\endgroup$
    – Em.
    Commented Jan 27, 2016 at 22:53
  • 1
    $\begingroup$ For part (b), equivalently worded, what is the probability that in the first $9$ cars, you use exactly one or exactly zero of the spare spark plugs? $\endgroup$
    – JMoravitz
    Commented Jan 27, 2016 at 22:54
  • $\begingroup$ In other words, you can avoid doing a summation if you ask yourself that question. $\endgroup$
    – Em.
    Commented Jan 27, 2016 at 22:58
  • $\begingroup$ thanks guys I am editing my post now please refresh it in a few minutes to check it. $\endgroup$
    – FACEIT
    Commented Jan 27, 2016 at 23:01

2 Answers 2


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*}



$X$ is the number of cars that can be repaired before both sparkplugs are used.

So, if $X=k$ then : $k-1$ cars have been repaired without one of them needing a change (in some order), and then one more car needed a change (the last car).   Can you evaluate the probability, $\mathsf P(X{=}k)$?

Do you recognise what type of probability distribution is this, or otherwise can you find the mean and variance?


Find: $\mathsf P(X > 10)$, which is the probability that $10$ cars can be repaired and among them there are either $1$ or $0$ changes required.  

Hint: the number of changes required among 10 repairs will be binomially distributed.

  • $\begingroup$ I am having troubles on part b, I don't know what to use as the probability. I let $Y$ be the no. of cars which require 1 or 0 plugs to be used, then we have $Y$ distrubted binomially with parameter $10$ and probability I think $1-0.05$ but I don't believe this yields a correct solution $\endgroup$
    – FACEIT
    Commented Jan 27, 2016 at 23:24
  • $\begingroup$ @Faceit Let $Y$ be the number of repairs which require a plug to be used. $\;$ So $Y\sim\mathcal{Bin}(10, 0.05)$. $\;$ Now evaluate the probability that less than two changes are required: $\quad\mathsf P(Y{=}0\cup Y{=}1)$ $\endgroup$ Commented Jan 27, 2016 at 23:36

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