Probability exercise (exponential variable)

Question

Can you help me to solve this exercise? I do not understand how to solve some points and my exam is coming soon. I have solved some items but I am in doubt about them

It asks:

A cell phone manufacturer offers warranty for $9$ months with a free replacement if the device was defective. The manufacturer estimates that each of the unit has an average life that corresponds to and exponential variable with mean equal to $24$ months

a) What probability of defective devicees will be replaced?

b) What is the density function for duration of devices that won't have any failure before $8$ months

c) Derek is a VIP member, he has infinite warranty. He can replace his phone every time is fails. If his first cellphone lasts more than a year, what is that probabilkity that $3$ phones won't cover $5$ year?

a)(NOT SURE)

$f(x) = \frac 1{24} e^{-x/24}$ , $x > 0$

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$P( x \leq 9) = \frac 1{24} \int_0^9 e^{-x/24}\text dx$

$u= \frac x{24}$

$\text du = \frac{\text dx}{24} $

\begin{align}\frac 1{24} \int e^{-x/24} \text dx &= \int e^{-u} \text du \\ \\&= -e^{-u} \\&= -e^{-x/24} \end{align}

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\begin{align}F(x) &= -e^{-x/24}\\ F(9) &= -e^{-9/24}\\ F(0) &= - e^0 \\ &= -1 \\ &\Downarrow\\ F(9)-F(0) &= 1- e^{-9/24} \\ &= 0.3127 \end{align}

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b)(NOT SURE)

$$P( x < 8) = \frac 1{24} \int_0^8 e^{-x/24}\text dx = 0.2835$$

Poisson with parameter $\lambda = 0.2835$ (The distribution of duration devices with failute time $< 8$ months)

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c)I have no idea

Answer 1

1. Seems right.

2. You have to find the conditional density $f_{X|X>8}(x)$, thus by definition $$ f_{X|X>8}(x) = \frac{f_X(x)I\{X>8\}}{P(X>8)} = \frac{\lambda e^{-\lambda x}I\{X>8\}}{e^{-\lambda 8}} = \lambda e^{-\lambda(x-8)}, \quad x\ge8. $$ Namely, it is a shifted exponential distribution.

3. Using the memoreyless property, we have $4$ years and maximum $3$ phones, namely you are asking what is the probability that a life duration of $3$ i.i.d phones will be less then $4$, i.e., $$ Y_3 = X_1 + X_2 + X_3 \sim \mathcal{G}(\lambda, 3), $$ hence $$ P(Y_3 \le 4) = \int_{0}^4 \frac{ \lambda ^3 }{2!}e^{-\lambda y}y^2dy = P(N(4)>3) = 1- \sum _{n=0}^3 \frac{e^{-4\lambda}(4\lambda)^n}{n!}. $$ Where $N(4) \sim Poiss (4 \lambda)$.