Binomial distribution problem with inconsistent probability of occurrence | 0.4  | 0.5  | 0.3  | 0.4  | 0.5  |
| 0.4  | 0.5  | 0.3  | 0.4  | 0.5  |
| 0.4  | 0.5  | 0.3  | 0.4  | 0.5  |
| 0.4  | 0.5  | 0.3  | 0.4  | 0.5  |

In each cell of the matrix is the winning probability.
What is the probability that more than 4 cells will win if all cells are randomized at the same time.
Because each lattice has a different probability, I can't solve it with a binomial distribution, and then I can't figure out what I should do.

There are $4$, resp. $8$ and $8$ cells with probability $0.3$, resp. $0.4$ and $0.5$.
The number of successes is:
$$X=X_1+X_2+X_3 \tag{1}$$
where $X_1 \sim Bin(4,0.3), \ \ X_2 \sim Bin(8,0.4), \ \ X_3 \sim Bin(8,0.5), \ \ $
The PGF of a Bin(n,p) distribution is $E(s^X)=(q+ps)^n$.
Therefore, as a consequence of (1), due to the independence of the $X_i$, the PGF of $X$ is $E(s^X)=E(s^{X_1})E(s^{X_2})E(s^{X_3})$, i.e.,
$$(0.7+0.3s)^4(0.6+0.4s)^8(0.5+0.5s)^8=p_0+p_1s+p_2s^2+\cdots p_{20}s^{20}\tag{2}$$
where $p_k:=P(X=k)$.
These coefficients can be obtained using Wolfram Alpha,

giving the final answer:
$$P(X>4)=1-P(X \le 4)=1-\sum_{k=0}^4 p_k \approx 1-0.0332812=0.9667188$$
Remark: $E(X)=4 \times 0.3 + 8 \times 0.4 + 7 \times 0.5=8.4$, this result being concordant with the fact that the most important values are $p_8$ and $p_{9}$.
