Question

Let \(A_{5\times5}\) be a matrix such that each of its elements follows \(Bernoulli(p=0.50)\) distribution independently.

The probability that the row-sum of the second row and the column-sum of the third column are both equal to 3 is __________. (Rounded off to two decimal places)

Hint:

Split the row-2 sum and column-3 sum around the one cell they share, \(A_{23}\). Condition on \(A_{23}\) being 0 or 1, then use \(Binomial(4,0.5)\) for the remaining four cells in each.

Answer 1

Correct Answer: 0.1

Step 1: Identify the cells involved.
The row-sum of the second row is the sum of the 5 entries \(A_{21}, A_{22}, A_{23}, A_{24}, A_{25}\).
The column-sum of the third column is the sum of the 5 entries \(A_{13}, A_{23}, A_{33}, A_{43}, A_{53}\).
Both sums share exactly one common cell, \(A_{23}\), which sits at the intersection of row 2 and column 3.
So the row-sum and the column-sum are not independent, they are linked through this one shared entry.

Step 2: Split the row and column into the shared cell and the rest.
Let \(X = A_{23}\), a single \(Bernoulli(0.5)\) variable.
Let \(R\) be the sum of the other 4 cells in row 2, that is \(A_{21}+A_{22}+A_{24}+A_{25}\).
Let \(C\) be the sum of the other 4 cells in column 3, that is \(A_{13}+A_{33}+A_{43}+A_{53}\).
Since all 25 entries are independent, \(X\), \(R\) and \(C\) are three independent random variables, and both \(R\) and \(C\) follow \(Binomial(4, 0.5)\).
Row-sum \(= R + X\) and column-sum \(= C + X\).
We need \(P(R+X=3 \text{ and } C+X=3)\).

Step 3: Write the Binomial(4, 0.5) probabilities.
For \(k = 0,1,2,3,4\), \(P(Binomial(4,0.5)=k) = \binom{4}{k}\left(\frac{1}{2}\right)^4 = \frac{\binom{4}{k}}{16}\).
\[ P(0)=\frac{1}{16}, \quad P(1)=\frac{4}{16}, \quad P(2)=\frac{6}{16}, \quad P(3)=\frac{4}{16}, \quad P(4)=\frac{1}{16} \]

Step 4: Condition on the value of the shared cell \(X\).
Case \(X=0\) (probability 0.5): we need \(R=3\) and \(C=3\).
\[ P(R=3)\cdot P(C=3) = \frac{4}{16}\cdot\frac{4}{16} = \frac{16}{256} = 0.0625 \]
Contribution: \(0.5 \times 0.0625 = 0.03125\)
Case \(X=1\) (probability 0.5): we need \(R=2\) and \(C=2\).
\[ P(R=2)\cdot P(C=2) = \frac{6}{16}\cdot\frac{6}{16} = \frac{36}{256} = 0.140625 \]
Contribution: \(0.5 \times 0.140625 = 0.0703125\)

Step 5: Add the two cases.
\[ P = 0.03125 + 0.0703125 = 0.1015625 \]
Rounded to two decimal places, this is 0.10.

Final Answer:
The probability that the row-sum of row 2 and the column-sum of column 3 are both equal to 3 is 0.10. \[ \boxed{0.10} \]