Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival time may be modeled by the random variable T, such that f(T = t) = {3/5 (5/t)^4 , t ≥ 5
0, otherwise
If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?
А. 62%
B. 73%
C. 88%
D. 91%
E. 96%

D. 91%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

In which

P(B|A) is the probability of event B happening, given that A happened.

is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

Simplifying:

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

Integral of is

Then

Applying the limits, by the Fundamental Theorem of Calculus:

At ,

At ,

Then

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

We already have the integral, so just apply the limits:

At ,

At ,

Then

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

use an equation of ratios to determine the amount of gas needed to fill the tank.

    0.2             0.8

  =    

288.75             x

                                                              231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

                                                                0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal         $3

* = $3465   (answer)

      1           1 gal

hope this you