Don't keep us apart!

Consider any 11 marks alone the rod, 9 of which indicate the break points and the other 2 being the marked points. Then of the $\dbinom{11}{9} = 55$ ways these marks can be ordered, $10$ have the 2 marked points being adjacent. As there is a one-to-one correspondence between these orderings and the outcomes specified by the question at hand, the desired probability is

$p = \dfrac{10}{55} = \dfrac{2}{11} \approx 0.182$ .

In general, with 2 points on a rod broken into $n$ pieces, the desired probability is $\dfrac{2}{n + 1}$ .

There are two sources of randomness: (1) breaking the rod into $n=10$ pieces and (2) choosing two marked points independently and uniformly at random. We will analyze the entire process in this sequence. Without any loss of generality, assume that the whole length of the rod is $1$ . If the lengths of the broken pieces are $\mathbf{x}=\{x_1,x_2, \ldots, x_n\}$ , then the conditional probability that the two marked points lie on the same piece is simply $p_{|\mathbf{x}}=\sum_i x_i^2,$ which follows from the fact that the points are chosen independently. Now to obtain the unconditional probability, we simply take the expectation of the above expression with respect to $\mathbf{x}$ . $p = \mathbb{E}\big(\sum_i x_i^2 \big) \stackrel{(a)}{=} \sum_i \mathbb{E}x_i^2 \stackrel{(b)}{=} n \mathbb{E}x_1^2,$ where (a) follows from linearity of expectation and (b) follows from the fact that the random variables $\{x_i\}_{i=1}^{n}$ have identical distributions. (note that, they are not independent- they sum to $1$ !). All that now remains is to find the expectation of $x_1^2$ . Now observe that $x_1$ will be more than $y$ iff all $n-1$ boundary points of the broken pieces lie outside the initial $y$ length of the rod, i.e., $\mathbb{P}(x_1 >y)= (1-y)^{n-1}.$ Thus, $\mathbb{E}(x_1^2)=\int_{0}^{1}\mathbb{P}(x_1^2 >y) dy = \int_{0}^{1} (1-\sqrt{y})^{n-1} dy = \frac{2}{n(n+1)}.$ Hence, $p= \frac{2}{n+1}.$

Alternatively, assume that the rod has length $1$ . If the rod is marked at points $0 \le x,y \le 1$ , then the $n-1$ break points that split the rod up into $n$ pieces must all lie outside the interval of length $|x-y|$ between $x$ and $y$ . Thus the desired probability is $\begin{aligned} \int_0^1 \int_0^1 \big(1 - |x-y|\big)^{n-1}\,dx\,dy & = 2\int_0^1\,dy \int_0^y (1 - y + x)^{n-1}\,dx \; = \; 2\int_0^1\,dy \int_0^y (1-x)^{n-1}\,dx \\ & = \tfrac{2}{n}\int_0^1\big(1 - (1-y)^n\big)\,dy \; =\; \tfrac{2}{n} \times \tfrac{n}{n+1} \; = \; \tfrac{2}{n+1} \end{aligned}$ In this case, the probability is $\tfrac{2}{11} = 0.1818\ldots$ .

By Trial and Error method, one can understand that there are ${n \choose 2}$ ways for the 2 points to be in different pieces where n is number of pieces.

So, for 10 pieces there are ${10 \choose 2}$ = 45 ways to arrange them in 10 different pieces of the rod .

And for the 2 marks to lie on the same piece there are 10 ways for total 10 pieces.

Total Number of ways (Possible Outcomes) = 45 + 10 = 55

P = Number of Favorable Outcomes / Total Number of Possible Outcomes

$P = 10/55 ≈ 0.182$

The probability is = the number of desired outcomes / total possible outcomes

There are 10 divisions. We want both dots to be on any single one. This desired outcome can happen on any of the 10 divisions. So number of desired outcomes is 10.

Total possible outcomes is the tricky part. Naming all divisions alphabetically starting from A we have ABCDEFGHIJ So arrangements of dots considering fist division can be AA, AB, AC, AD, AE, AF, AG, AH, AI and AJ. So thats 10 outcomes. We might conclude that so for all 10 divisions it should be 10 x 10= 100. But let's consider the second division B. The AB variation is already included in the first 10 outcomes. Hence we need to consider 9. For division C, both AC and BC must have been accounted for, so we consider 8. So on and so forth the total possible outcomes = 10 + 9 + 8 + 7 + 6 + 5 + 4 +3+2+1 = 55 Probability is 10/55 ≈ 0.182

Let's assume you have already divided the rod into 10 pieces. Now for both dots to be on the same piece, Favourable methods are 10C1 and total ways to mark two dots is 10C1 + 10C2. Thus Probability is 10/55