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If the number of minutes it takes a service station attendant to balance a tire is a random variable having an exponential distribution with the parameter $\lambda = 0.2$, what are the probabilities that the attendant will take:

a) Less than $8$ minutes to balance $2$ tires

b) Less than $12$ minutes to balance $3$ tires

The $\lambda$ parameter confuses me, usually the parameter is $\theta$ so I don't know if that makes a difference, also I don't know how to figure out more than $1$ tire.

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The λ parameter confuses me,

An exponential distribution with parameter $\lambda$ is the one with expectation $\lambda^{-1}$.

PDF: $f_X(x)=\lambda\,\mathsf e^{-\lambda x}\,\mathbf 1_{x\in[0;\infty)}$

CDF: $F_X(x)=(1-\mathsf e^{-\lambda x})~\mathbf 1_{x\in[0;\infty)}$


I don't know how to figure out more than 1 tire.

Use convolution.

$$\mathsf P(X_1+X_2\leq z)=\int_0^z\int_0^{z-x} f_{X_1}(x)f_{X_2}(y)~\mathsf d y~\mathsf d x$$

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I think the process your are talking about is a Poisson counting process. The time elapsed until the attendant is able to balance a new tire is $1/\lambda$ that is every 5 minutes. The expected number of tires he is able to balance in a given time duration $t$ (expressed in minutes) is $E[N]=\lambda t$. This number is a random variable of probability density,

$p(N, t)= \frac{(\lambda t)^N}{N!}\exp(-\lambda t)$

I would say that the probabilities you want to determine are expressed by,

$P(N,t\leq t_{\rm max})= \frac{\int_0^{t_{\rm max}} \frac{(\lambda t)^N}{N!} \exp(-\lambda t) dt}{\int_0^\infty \frac{(\lambda t)^N}{N!}\exp(-\lambda t) dt}$

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