TITLE: Reading Binomial Distribution Tables
QUESTION [0 upvotes]: 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.

REPLY [1 votes]: [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. 

I changed the table - does this help? I still don't understand what it means? Is it saying the probability of getting at least one in ten trials if the probability is 10% is 100%? 

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.
$$\begin{align}\mathsf P(X\leqslant 5;p=0.1)&=0.999\,853\,097\,4\\[1ex]&\approx1.000\end{align}$$
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