Suppose I have a 5x5 grid that every cell can contain one of two numbers $\ 0, 1 $ . What is the probability that at least in one row, the sum of numbers in that row will be $\ 3 $ .
I'm trying to follow the same logic as instructed in this question
So the probability of row $\ i $ to have a sum of $\ 3 $ is $\ \frac{{5 \choose 3}}{2^5} $ but this is not very helpful. so how do i construct it such that i'll be given the probability of all rows sums to be different than $\ 3 $ ?
Since the cell values are independent, the probability that a given row has a sum of $3$ is $${\small{\binom{5}{3}}}\left({\small{\frac{1}{2}}}\right)^5={\small{\frac{5}{16}}}$$ Hence the probability that a given row does not have a sum of $3$ is $$1-{\small{\frac{5}{16}}}={\small{\frac{11}{16}}}$$ Since the cells values are independent, so are the row sums, hence the probability that for all $5$ rows, the row sum is not equal to $3$ is $$ \left( {\small{\frac{11}{16}}} \right)^5 = {\small{\frac{161051}{1048576}}} $$ Hence the probability that at least one row has a sum of $3$ is $$1-{\small{\frac{161051}{1048576}}}={\small{\frac{887525}{1048576}}}$$