Q. 5.15 - A First Course in Probability

Question

The number of years that a washing machine functions is a random variable whose hazard rate function is given by

$\lambda \left(t\right)=\left\{\begin{array}{l}.2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<t<2\\ .2+.3(t-2)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\le t<5\\ 1.1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}t>5\end{array}\right.$

(a)What is the probability that the machine will still be working $6$ years after being purchased?

(b) If it is still working $6$ years after being purchased, what is the conditional probability that it will fail within the next

$2$ years?

Step-by-Step Solution

Answer

(A) The probability of machine when it is  working $6$ years after purchased is $P(X>6)=e^{-3.45}\approx 0.0317$

(B) The conditional probability that it will fail within the next $6$ years $P(X>8)=e^{-5.65}\approx 0.003517$

Step :1 A random variables

Create a random variable. The number $X$ represents the machine's lifetime. It is assumed that the provided hazard function defines $X$.

We'll use the relationship between the hazard rate and the cumulative function provided by

$F\left(t\right)=1-\exp \left(-{\int }_0^t \lambda \left(s\right)ds\right)$

Step : 2 The probability that the machine will still be working 6 years after being purchased (part a)

The task at hand is to locate $P(X>6)$. We can deduce from the above link that

$$\begin{gathered} P(X>6)=1-P(X\le 6) \\ =1-F\left(6\right) \\ =\exp \left(-{\int }_0^6 \lambda \left(s\right)ds\right) \end{gathered}$$

Let's calculate ${\int }_0^6 \lambda \left(s\right)ds$. We have that

${\int }_0^6 \lambda \left(s\right)ds={\int }_0^2 0.2ds+{\int }_2^5 (0.2+0.3(s-2\left)\right)ds+{\int }_5^6 1.1ds$

$$\begin{gathered} =0.4+1.95+1.1 \\ =3.45 \end{gathered}$$

So we have that 

$P(X>6)=e^{-3.45}\approx 0.0317$

Step :3 Conditional probability (part b)

We are required to find $P(X\le 8\mid X>6)\text{.}$ . It is equal to

$P(X\le 8\mid X>6)=\frac{P(6<X<8)}{P(X>6)}$

The probability in the denominator is equal to

Probability $P(X>8)$  is calculated similarly as in (a). We have that

${\int }_0^8 \lambda \left(s\right)ds={\int }_0^2 0.2ds+{\int }_2^5 (0.2+0.3(s-2\left)\right)ds+{\int }_5^8 1.1ds$

$$\begin{gathered} =0.4+1.95+3.3 \\ =5.65 \end{gathered}$$

So,

$P(X>8)=e^{-5.65}\approx 0.003517$