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A man leaves his house at the same time every morning and the time taken to journey to work has the following probability density function:

  • less than 30 minutes, zero;
  • between 30 minutes and 60 minutes, uniform with density $k$;
  • between 60 minutes and 70 minutes, uniform with density $2k$;
  • more than 70 minutes, zero.

What is the probability that on one particular day he arrives at work later than on the previous day but not more than 5 minutes later?

My Solution: $$f(x) = \begin{cases} k; & 30 <x< 60 \\ 2k ; & 60 <x< 70 \\ 0 ; & o.w. \end{cases}$$ Using probability axiom, the value of $k$ will be 1/50. Since, if he was 30 minutes late previous day then today he can be later not more than 35 minutes. So, the possible chances that he is later on the next is from 35 to 70 minutes. So, $$ P(E) = P ( 35 <X< 70 )$$ where E = {one particular day he arrives at work later than on the previous day but not more than 5 minutes later}. After calculations, $P (E) = 9/10$. Correct me if I am wrong.

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  • $\begingroup$ My Solution : f(x) = k ; 30 <x< 60 , 2k ;60 <x< 70 , 0 ;o.w. using probability axiom, The value of "k" will be 1/50. Since, if he was 30 minutes late previous day then today he can be later not more than 35 minutes. So, the possible chances that he is later on the next is from 35 to 70 minutes. So, P(one particular day he arrives at work later than on the previous day but not more than 5 minutes later) = P ( 35 <X< 70 ) After calculations, P ( 35 <X< 70 ) = 9/10 . Correct me if I am wrong. $\endgroup$ – Abhay singh Apr 22 at 17:22
  • $\begingroup$ Your work should be in the post and not in comments. $\endgroup$ – StubbornAtom Apr 22 at 20:24
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The probability in question is: $$\begin{align} &\int_{-\infty}^\infty\rho(t)dt\int_t^{t+\Delta t}\rho(t')dt'\\ &=\int_{30}^{55}(k)[k\cdot5] dt+\int_{55}^{60}(k)[k(60-t)+2k(t-55)]dt\\ &+\int_{60}^{65}(2k)[2k\cdot 5]dt+\int_{65}^{70}(2k)[2k(70-t)]dt\\ &=k^2(125+\frac{75}2+100+50)=\frac18, \end{align} $$ where $\rho(t)$ is the probability density, and $\Delta t=5$.

Explanation: The integral limits denote the time range of the arrival on the previous day and the number in parentheses - the corresponding probability density (to be multiplied with $𝑑𝑡$ to obtain the probability to arrive between $𝑡$ and $𝑡+𝑑𝑡$). The number in brackets denotes the probability to arrive on the next day between $𝑡$ and $𝑡+\Delta t$ (i.e. $\int_t^{t+\Delta t}\rho(t')dt'$).

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  • $\begingroup$ Please explain your steps!!! I am confused!! $\endgroup$ – Abhay singh Apr 22 at 19:29
  • $\begingroup$ [k(60-t) +2k(t-55)] what it denotes? $\endgroup$ – Abhay singh Apr 22 at 19:42
  • $\begingroup$ Why you took limits 30 to 55 and then 55 to 60 ? $\endgroup$ – Abhay singh Apr 22 at 19:43
  • $\begingroup$ Okay thanks sir!! $\endgroup$ – Abhay singh Apr 22 at 19:46
  • $\begingroup$ @Abhaysingh I added the explanation into the body of the answer. $\endgroup$ – user Apr 22 at 19:58

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