Time-based probability question Two adult male baboons are introduced to the same $50$ ft. square cage.  Male A looks at Male B for a total of $5$ hours in the first ($24$ hour) day, and Male B looks at Male A for a total of $3$ hours in the same first day.  If the two baboons behave independently of each other, what is the probability that they will look at each other at the same time (at least once) on that first day?

My thinking: I think this can be solved geometrically, but I am having trouble interpreting it. Can I interpret it to $0 &lt; B &lt; 5$ and $0 &lt; A &lt; 3$ and $A-B &gt; 0$?

Another solution I had for this question is that: Two baboons only look together if A's $5$ hour is happening within the $3$ hours of B looking at A. So the probability would be $\frac{3}{5}\cdot\frac{5}{24}=\frac{1}{8}$? I think it is wrong though.
Decomposing the day into $n$ equal timeslots with $n$ large and assuming each animal chooses randomly uniformly and independently the necessary number of timeslots, one gets a probability of no eye contact approximately equal to $\mathrm e^{-n/10}.$


Let $u=5/24$ and $v=8/24$, then Male A looks at Male B during $a$ timeslots, with $a=un$, and Male B looks at Male A during $b$ timeslots, with $b=vn$. Fix the $a$ timeslots used by Male A. No eye contact happens if Male B uses $b$ distinct timeslots taken from the $n-a$ remaining "free" timeslots. Consider that Male B chooses his $b$ timeslots sequentially. The first timeslot Male B chooses is free with probability $(n-a)/n$. Conditionally on this, the second timeslot Male B chooses is free with probability $(n-a-1)/(n-1)$, and so on until timeslot number $b$, which is free with conditional probability $(n-a-b+1)/(n-b+1)$. Thus, the exact probability of no eye contact is $$p=\prod_{i=0}^{b-1}\frac{n-a-i}{n-i}=\frac{(n-a)!\,(n-b)!}{n!\,(n-a-b)!}.$$ When $u$ and $v$ are fixed and $n\to\infty$ with $a=un$ and $b=vn$, $$\frac{\log p}n\to-\theta=(1-u)\log(1-u)+(1-v)\log(1-v)-(1-u-v)\log(1-u-v),$$ hence one can approximate $p$ as $$p\approx\mathrm e^{-\theta n}.$$
If $u=5/24$ and $v=8/24$, $\theta\approx.09768249$. For a decomposition into hours, $n=24$ hence $p\approx9.6\%$. For a decomposition into minutes, $n=1440$ hence $p\approx10^{-61}$. 
For a decomposition into seconds, $n=86,400$ hence $p\approx10^{-3665}$.



Having said that (which solves the problem as interpreted by the OP), my impression is that the real question assumes some continuous periods of 5 hours, respectively 3 hours, in the day, and asks for the probability of no eye contact, that is, that the time intervals corresponding to these periods do not intersect. 

This happens if and only if the midpoints $X$ and $Y$ of the intervals are at distance more than $4$, $X$ being uniform on $(2.5,21,5)$ and $Y$ uniform on $(1.5,22.5)$. Equivalently, one can consider $X$ uniform on $(0,19)$ and $Y$ uniform on $(0,21)$. Then $p$ is the normalized sum of the area of the two triangles $X\gt Y+4$ and $Y\gt X+4$, that is, $$p=\frac{\frac1217^2+\frac1215^2}{19\cdot21}=\frac{257}{399}\approx64\%.$$
When I revisit this question:

Let event A be that the male baboon A looking at the male baboon B. P(A)=5/24

Let event B be that the male baboon B looking at the male baboon A. P(B)=3/24

So P(A and B)=5/24*3/24=5/192