jmc

First we find the total number of elements in the first $15$ rows. The first row of Pascal's Triangle has one element, the second row has two, and so on. The first $15$ rows thus have $1+2+\cdots +15$ elements. Instead of manually adding the summands, we can find the sum by multiplying the average of the first and last term $\frac{1+15}{2}$ by the number of terms, $15$. The sum is $\frac{16}{2}\cdot 15=8\cdot 15=120$, so there are $120$ elements. Now we find the number of ones in the first $15$ rows. Each row except the first has two ones, and the first row only has one. So there are $14\cdot 2+1=29$ ones. With $29$ ones among the $120$ possible elements we could choose, the probability is $\boxed{\frac{29}{120}}$.