# Exponential distribution problem finding probabilities

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.

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.

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

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,

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