A square is divided into n^2 equal squares. A ball is thrown…
A square is divided into $n^2$ equal squares. A ball is thrown into it; the probability that it falls into the small square in the $i$-th horizontal and $j$-th vertical strip is $p_{ij}$, with $\sum_{i,j=1}^np_{ij}=1$. Determine the probability that the ball falls into the $k$-th horizontal strip.
Explanation of Method
The $k$-th horizontal strip consists of the $n$ small squares $(k,1),(k,2),\dots,(k,n)$. The ball falls into at most one small square, so these events are mutually exclusive and their probabilities add up. The answer is the sum of the $k$-th row of the table $(p_{ij})$ — a "marginal" probability.
Step 1: The strip as a sum of cells
Let $A_{ij}$ = "the ball falls into cell $(i,j)$", $P(A_{ij})=p_{ij}$. The $k$-th horizontal strip is
\[ H_k = A_{k1}+A_{k2}+\dots+A_{kn}, \]
and different cells are mutually exclusive.
Step 2: Addition rule
\[ \boxed{P(H_k)=\sum_{j=1}^{n}p_{kj}=p_{k1}+p_{k2}+\dots+p_{kn}.} \]
Similarly, the $l$-th vertical strip has probability $\sum_{i=1}^np_{il}$ (the sum of a column).
Step 3: Example with $n=3$
| column 1 | column 2 | column 3 | row sum | |
|---|---|---|---|---|
| row 1 | $0.05$ | $0.10$ | $0.05$ | $0.20$ |
| row 2 | $0.10$ | $0.30$ | $0.10$ | $0.50$ |
| row 3 | $0.05$ | $0.15$ | $0.10$ | $0.30$ |
The probability of the second horizontal strip is $0.1+0.3+0.1=0.5$. The row sums add up to $1$, as they must.
Step 4: Meaning
This is how a joint distribution of two quantities (row number, column number) gives the distribution of one of them: add over all values of the other.
Result
| Answer | |
|---|---|
| Probability of the $k$-th horizontal strip | $\sum_{j=1}^np_{kj}$ |