4

I'm having trouble solving this exercise about probability, as I'm new to the subject. It says:

There is a bus supposed to arrive at 8:00. The probability of it to set out with $m$ minutes of delay is $p_m = 3^{-m} (m = 1; 2; 3; ... ;m \neq 0)$.
Due to regulatory issues, the set out is never advanced (i.e. it never sets out before 8:00).

1) What's the probability of the bus setting out at 8:00?
2) Prove that the probability of the bus to set out at 8:10 is double the probability that of after 8:10.

Here is what I think about it, but I have no answer to check with:
I assume all possible values go from 8:00 to infinity, so if I add all those values the probability should be 1. Now, $3^{-m}$ looks like a series I know, and I know it converges as 1/3 is less than 1, so the sum of all terms is $1*\frac{1}{1-1/3}=\frac{3}{2}$, but as that formula is when n starts at 0, then I have to do $\frac{3}{2}-3^0$, which is 0.5.
Then, I guess that 1-0.5 is the remaining probability, that is, of the bus setting out exactly at 8:00. Is what I've done correct? Because I don't think we're mixing calculus with probabability in the course.

For 2), my guess is calculate $3^{-10}$, and I have to prove that $3^{-10} = 2*\sum_{m=11}^\infty 3^{-m}$. Is it correct? What I do next is calculate the sum, which I don't really know how to do, so I do $3/2 -\sum_{m=0}^{10} 3^{-m}$, put that in the calculator and I get that it's true.

So, to wrap it all up... Have I done everything ok? Because I don't have the results. I suppose the second one is correct because I've proven it, but I've no idea about the first one, whether I'm meant to do that or there is another way.

I hope you can help me. Thank you!

Key Flex
  • 9,475
  • 7
  • 17
  • 34

1 Answers1

1

You have (1) correct. There is no calculus involved, just geometric series and probability.

From geometric series we indeed have: $\mathsf P(m\neq0)~{=\sum_{m=1}^\infty 3^{-m}\\=\frac 12\qquad\checkmark}$

From the law of complementary probability: $\mathsf P(m=0)~{=1-\mathsf P(m\neq 0)\\=\tfrac12\qquad\checkmark}$

For (b) note that $\mathsf P(m>10)~{=\sum_{m=11}^\infty 3^{-m}\\=3^{-10}\sum_{n=1}^\infty 3^{-n}\\~~\vdots}$

Graham Kemp
  • 129,094
  • Is that a property or how did you do that in (b)? –  Jul 17 '18 at 14:52
  • 1
    Distribution of common factor, $\sum_{m=11}^\infty 3^{-m}=3^{-10}\sum_{m=11}^\infty 3^{10-m}$, and a change of bound variable, $\sum_{m=11}^\infty 3^{10-m}=\sum_{n+10=11}^\infty 3^{10-(n+10)}$. @Esteban – Graham Kemp Jul 17 '18 at 15:46
  • Oh, I understand. Thank you! –  Jul 17 '18 at 15:53