Probability of a Gamma distribution using binomial probabilities

Question

Suppose we have a distribution $Y\sim Gamma(\alpha=4, \beta=7)$

Given $F(Y)=\sum_{i=\alpha}^{n}{n \choose i}y^i(1-y)^{n-i}$, where $n=\alpha+\beta+1$, I need to find $P(Y \leq .7)=F(.7)$ using binomial tables.

By substituting the parameters, this gives $F(.7)=\sum_{i=4}^{10}{10 \choose i}.7^i(.3)^{10-i}=.989$

Can someone please explain to me how to get $.989$ from the sum using this table? I know it corresponds to $a=6$ and $p=.4$ in the $n=10$ part, but I do not understand why we must look at this cell with $p=.3$ instead of $p=.7$

Thanks!

Answer 1

We make a few observations $$P(Y\geq 4)= \sum_{i = 4}^{10} p^i (1-p)^{10-i} = 1-P(Y\leq 3) = 1-\sum_{i=0}^3p^i(1-p)^{10-i}$$

The bounds on your table are $0$ to $a$. Hence, we look up $n=10, a = 3, p = .7$ and get $$\sum_{i=0}^3p^i(1-p)^{10-i}\approx .011.$$ So our approximation is $$1-P(Y\leq 3) = 1-\sum_{i=0}^3p^i(1-p)^{10-i}\approx 1-.011 =.989$$