Weighted Coin, Probability Question

Let $X$ be the number of heads in 40 flips. The head probability is $p=0.1429$ and the tail probability is $q=1-p=0.8571$. Then $X\sim \mathrm{Binomial}(40,p)$ and
$$ \mathbb{P}(X=k)=\binom{40}{k}p^{k}q^{40-k}.$$

Compute the small-$k$ terms:

$$\mathbb{P}(X=0)=q^{40}=(0.8571)^{40}\approx 0.002095.$$

$$\mathbb{P}(X=1)=\binom{40}{1}p\,q^{39}=40(0.1429)\,(0.8571)^{39}\approx 0.013974.$$

$$\mathbb{P}(X=2)=\binom{40}{2}p^{2}q^{38}=780(0.1429)^2\,(0.8571)^{38}\approx 0.045431.$$

$$\mathbb{P}(X=3)=\binom{40}{3}p^{3}q^{37}=9880(0.1429)^3\,(0.8571)^{37}\approx 0.095944.$$

Add them:
$$\mathbb{P}(X\le 3)\approx 0.002095+0.013974+0.045431+0.095944 \approx 0.157445.$$

Therefore
$$\mathbb{P}(X\ge 4)=1-\mathbb{P}(X\le 3)\approx 1-0.157445=\boxed{0.8426}.$$

Using the exact fraction $p=\tfrac{1}{7}=0.1429$ gives $\mathbb{P}(X\ge 4)\approx 0.8424$ (very close).