# In I=$[0,1]$, probability B Leaves right after A Arrives

Sorry if this is to simple; I have not done any Math in a while.

The scenario is , we have two participants/Runners: A,B both headed

randomly towards the same place P, and staying in P for a random interval

then heading out. We want to know the probability of

B arriving right after A leaves.

So we normalize meeting at $$[0,1]$$ and A arrives at $$a_1$$ and stays

until $$a_2$$; and similar pair $$(b_1, b_2)$$ for B; $$0 \geq a_1,a_2,b_1,b_2 \leq 1$$.

So we want to compute the probability that $$a_2=b_1$$

Working on $$[0,1] \times [0,1]$$. The probability of the event $$a_2=b_1$$

is the measure of all rectangles with base $$(a_1,a_2)$$ and side $$(a_2,1)$$,

with total area $$(a_2- a_1)(1-a_2)$$. WE can compute the total area by the

double integral:

$$\int_0^1 (a_2-a_1)da_1 \int_0^1 (1-a_2)da_2$$

Getting 1/12 , which seems somewhat right intuitively. But I have been told

(non-rigorously) that the actual answer is 0. Can someone check and

criticque my work, please?

• It seems that A and B are continuous random variables. Then $P(A-B=c)=0$. The probabilities are positive only for inequalities. – callculus Oct 11 '18 at 17:29
• But what is wrong with my answer then? – gary Oct 11 '18 at 17:32
• @callculus: I am not saying I am not wrong, I would just like to know what false assumption I am making in my post. – gary Oct 11 '18 at 17:52