A subway has good service 70% of the time and runs less frequently 30% of the time because of signal problems. When there are signal problems, the amount of time in minutes that you have to wait at the platform is described by the pdf probability density function with signal problems = pT|SP(t) = .1e −.1t But when there is good service, the amount of time you have to wait at the platform is probability density function with good service = pT|Good(t) = .3e −.3t You arrive at the subway platform and you do not know if the train has signal problems or is running with good service, so there is a 30% chance the train is having signal problems. (a) What is the probability that you wait at least 1 minute if there is good service? (b) What is the probability that you wait at least 1 minute if there are signal problems? (c) After 1 minute of waiting on the platform, you decide to re-calculate the probability that there are signal problems conditioning on the fact that your wait will be at least 1 minute long (since you have already waited 1 minute). What is that new probability? (d) After 5 minutes of waiting, still no train. You re-calculate again. What is the new probability?

A subway has good service 70% of the time, that is, [tex]P(GS)=0.7[/tex] and a subway runs less frequently 30% of the time because of the signal problems, that is, [tex]P(SP)=0.3[/tex].

If there are signal problems, the amount of time T in minutes that have to wait at the platform is described by the probability density function given below:

[tex]P_{T|SP}(t)=0.1e^{0.1t}[/tex]

If there is good service, the amount of time T in minutes that have to wait at the platform is described the probability density function given below:

[tex]P_{T|GOOD}(t)=0.3e^{0.3t}[/tex]

(a)

The probability that you wait at least 1 minute if there is good service P(T ≥ 1| GS) is obtained as follows:

[tex]P(T\geq 1|GS)=\int\limits^{\infty}_1 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_1 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_1\\\\=-(0-e^{-0.3})\\=0.74[/tex]

(b)

The probability that you wait at least 1 minute if there is signal problems P(T ≥ 1| SP) is obtained as follows:

[tex]P(T\geq 1|SP)=\int\limits^{\infty}_1 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_1 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.3})]\\\\=-(e^{-0.1t})\limits^{\infty}_1\\\\=-(e^{\infty}-e^{-0.1})\\=-(0-0.904)\\=0.904[/tex]

(c)

After 1 minute of waiting on the platform, the train is having signal problems follows an

exponential distribution with parameter [tex]\lambda= 0.1[/tex]

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is obtained using the result given below:

[tex]P(SP|T\geq 1)=\frac{P(T\geq 1|SP)P(SP)}{P(T\geq 1)}[/tex]

[tex]P(T\geq 1|GS)=0.74, P(T\geq 1|SP)=0.904[/tex]

Now calculate the [tex]P(T \geq 1)[/tex] as follows:

[tex]P(T \geq 1)=P(T\geq 1|SP)P(SP)+P(T\geq 1|GS)P(GS)\\=(0.904)(0.3)+(0.74)(0.7)=0.7892[/tex]

The probability that the train is having signal problems based on the fact that will be at least 1 minute long is calculated as follows:

[tex]P(SP|T\geq 1)= \frac{0.904 \times 0.3}{0.7892} = 0.3436 [/tex]

Hence, the probability that the train is having signal problems based on the fact that will be at least 1 minute long is [tex]0.3436[/tex].

(d)

After 5 minutes of waiting on the platform, the train is having signal problems follows an exponential distribution with parameter [tex]\lambda= 0.1[/tex].

The probability that the train is having signal problems based on the fact that will be at least 5 minutes long is obtained using the result given below:

[tex]P(SP|T\geq 5)=\frac{P(T\geq 5|SP)P(SP)}{P(T\geq 5)}[/tex]

First, calculate the [tex]P(T\geq 5|SP)[/tex] as follows:

[tex]P(T\geq 5|SP)=\int\limits^{\infty}_5 {0.1e^{-0.1t}} \, dt\\\\=0.1\int\limits^{\infty}_5 {e^{-0.1t}} \, dt\\\\=0.1[(\frac{e^{-0.1t}}{-0.1})]\\\\=-(e^{-0.1t})\limits^{\infty}_5\\\\=-(e^{\infty}-e^{-0.5})\\=-(0-0.6065)\\=0.6065[/tex]

Now, calculate the [tex]P (T\geq5|GS )[/tex] as follows:

[tex]P(T\geq 5|GS)=\int\limits^{\infty}_5 {0.3e^{-0.3t}} \, dt\\\\=0.3\int\limits^{\infty}_5 {e^{-0.3t}} \, dt\\\\=0.3[(\frac{e^{-0.3t}}{-0.3})]\\\\=-(e^{-0.3t})\limits^{\infty}_5\\\\=-(0-e^{-1.5})\\=0.2231[/tex]

Now, calculate the [tex]P (T \geq 5)[/tex] as follows:

[tex]P(T \geq 5)=P(T\geq 5|SP)P(SP)+P(T\geq 5|GS)P(GS)\\=(0.6065)(0.3)+(0.2231)(0.7)=0.3381[/tex]

The probability that the train is having signal problems based on the fact that will be at least 5 minutes long is calculated as follows:

[tex]P(SP|T\geq 5)= \frac{0.6065 \times 0.3}{0.3381} = 0.5381 [/tex]

Hence, the probability that the train is having signal problems based on the fact that will be at least 1 minute long is [tex]0.5381[/tex].