Exam type problem Assume that the probability of being stopp

Exam type problem: Assume that the probability of being stopped by the police in a time interval of length T years is integral 0 to T 0.18 e^-0.18X dx Find the probability of being stopped next year not being stopped the next 10 years How many months will go by before your chance of being stopped is more than 50%

Solution

We are given that

0.18 * e^-0.18X (0 to T)

P(being stopped next year) = 0.18 * e^-0.18X    (0 to 1)

= 0.18 * [  e^-0.18X / -0.18 ]

= - [ e^-0.18*1 - e^-0.18*0 ]

= 1 - 0.8353 = 0.1647

P(not being stopped the next 10 years) = 1 - P(being stopped the next 10 years)

= 1 - 0.18 * e^-0.18X (T is from 0 to 10)

= 1 - [ 0.18 * [  e^-0.18X / -0.18 ] ]

=1 - ( - [ e^-0.18*10 - e^-0.18*0 ]

= 1 - [ -0.1653 + 1]

= 0.1653

How many months will go by before your chance of being stopped is more than 50%.

That mean we have given probability and we have to calculate Time.

0.18 * e^-0.18X = 0.5 (0 to T)

0.18 [ e^-0.18T / -0.18] = 0.5

- [  e^-0.18T - e^0.18*0] = 0.5 (0 to T)

- [ e^-0.18T - 1 ] = 0.5

1 - e^-0.18T = 0.5

e^-0.18T = 1 - 0.5

e^-0.18T = 0.5

take log to the base e on bothsides,

-0.18*T = log_e(0.5)

-0.18*T = -0.69315

T = -0.69315 / -0.18

T = 3.85

Approximately equal to 4.