If I have a $ 5\times5$ board that in each cell could have one of two numbers; $ 0, 1 $. What is the probability that there will be at least one row of only zeros?
So the sample space is $\ 2^{25} $ (?) and at first I tried to think of each event. Like $\ A_i $ will be the event that the row $\ i $ will be zeros but there are $5$ different events and it's getting to complicated to calculate it.
Any hints how could I make this problem easier?
On
Let $R_i$ denote the number of possibilities with only zero's in row $i$.
To be found is $2^{-25}|R_1\cup R_2\cup R_3\cup R_4\cup R_5|$ and with the principle of inclusion/exclusion and symmetry we find:$$|R_1\cup R_2\cup R_3\cup R_4\cup R_5|=$$$$5|R_1|-10|R_1\cap R_2|+10|R_1\cap R_2\cap R_3|-5|R_1\cap R_2\cap R_3\cap R_4|+|R_1\cap R_2\cap R_3\cap R_4\cap R_5|=$$$$5\cdot2^{20}-10\cdot2^{15}+10\cdot2^{10}-5\cdot2^5+1\cdot2^0$$
On
There is only $1$ out of $32$ possible ways to fill a row with all $0$'s
If follows that there are $31$ ways to fill a row without it having all $0$'s.
Hence, there are $31^5$ ways to fill all $5$ rows without any of them having all $0$'s
So, there are $32^5-31^5$ ways to fill the $5$ rows with at least of them having all $0$'s
Therefore, the probability of having at least one row with all $0$'s is:
$$\frac{32^5-31^5}{32^5}$$
You want that the probability that it's not the case that every row has at least one nonzero.
For any given row to have at least one nonzero, you want that it's not all zeroes. The probability of that's $1-(\frac12)^5$.
Then that the above is untrue for at least one row you want to take away from certainty, the probability it's true for every row:
$$1-\left(1-\left(\frac12\right)^5\right)^5=\frac{4925281}{33554432}$$