This is a binomial distribution table. I understand that what this means is that the probability of 2 successes in 6 trials with a probability of 20% is 98%. But I think I am misunderstanding something. How can the probability of 4 successes be 100% when the probability of success is only 10%? How can any of the probabilities be 100%? Is there some way that I am not reading this correctly?
EDIT: I see it has something to do with the fact that this shows the cumulative binomial probability, but I still don't understand how the probability is understood for those that are 1.
[edit: now using $X\sim\mathcal{Bin}(10,p)$, as per the new table]
The new table includes the column labels, making it clearer that it measures cummulative probability, and also it still rounds to three decimal places.
No, it is saying that, when the rate for success is $0.1$, then the probability for at most one success among ten trials is approximately $0.736$.
$$\begin{align}\mathsf P(X\leqslant 1; p{=}0.1) &=\sum_{k=0}^1\binom {10}k(0.1)^k(0.9)^{10-k}\\[1ex]&=\tbinom{10}0~0.1^0~0.9^{10}+\tbinom{10}1~0.1^1~0.9^9\\[1ex]&=0.348\,678\,440\,1+ 0.387\,420\,489\\[1ex]&=0.736\,098\,929\,1\\[1ex]&\approx. 0.736\end{align}$$
The values decrease as you move right along a row; because as the success rate increases then the probability for obtaining at most one success goes down (as instead you become more likely to obtain more than one success).
$$\begin{align}\mathsf P(X\leqslant 1;p=0.6)&=\tbinom{10}0~0.6^0~0.4^{10}+\tbinom{10}1~0.6^1~0.4^9\\[1ex]&=0.001\,677\,721\,6\\[1ex]&\approx0.002\end{align}$$
And likewise as you move down a column, the value of probability increases, and it may well eventually round to $1.000$ at three decimal places.
$$\begin{align}\mathsf P(X\leqslant 5;p=0.1)&=0.999\,853\,097\,4\\[1ex]&\approx1.000\end{align}$$