A5×5 each element following Bernoulli (P = 0.50) Dist. independently. The prob. that row sum of the second row and column sum of the third column are both equal to 3 is

The correct answer is 0.10.

Let $A_{5\times 5}$ be a 5 by 5 matrix where each element $a_{i,j}$ is an independent Bernoulli random variable with success probability $p=0.50$. This means each element can take the value 1 with probability 0.50, or 0 with probability 0.50.

We want to find the probability that the row sum of the second row (Row 2) is equal to 3, and the column sum of the third column (Column 3) is equal to 3, simultaneously.

Let $R$ be the sum of elements in Row 2:
$R=a_{2,1}+a_{2,2}+a_{2,3}+a_{2,4}+a_{2,5}$

Let $C$ be the sum of elements in Column 3:
$C=a_{1,3}+a_{2,3}+a_{3,3}+a_{4,3}+a_{5,3}$

Notice that these two sums share exactly one matrix element in common, which is the intersection element $a_{2,3}$. The remaining elements in Row 2 and Column 3 are all independent of each other. Let us analyze the probability by conditioning on the value of this shared element, $a_{2,3}$.

Case 1: $a_{2,3}=1$
This event occurs with probability $P(a_{2,3}=1)=0.50$.
If $a_{2,3}=1$, then for the sum of Row 2 to equal 3, the remaining 4 elements in Row 2 must sum to 2. These 4 elements are independent Bernoulli(0.50) variables, so their sum follows a Binomial distribution with parameters $n=4$ and $p=0.50$.
The probability that their sum is 2 is:
$P(\text{Row 2 sum}=3\mid a_{2,3}=1)=\left(\genfrac{}{}{0ex}{}{4}{2}\right){0.5}^2{0.5}^2=6\times 0.0625=0.375$
Similarly, for the Column 3 sum to equal 3, the remaining 4 elements in Column 3 must sum to 2. This also follows a Binomial distribution with $n=4$ and $p=0.5$:
$P(\text{Column 3 sum}=3\mid a_{2,3}=1)=\left(\genfrac{}{}{0ex}{}{4}{2}\right){0.5}^2{0.5}^2=0.375$
Since the remaining elements of Row 2 and Column 3 are disjoint and thus independent, the conditional joint probability for Case 1 is:
$P(R=3\cap C=3\mid a_{2,3}=1)=0.375\times 0.375=0.140625$

Case 2: $a_{2,3}=0$
This event occurs with probability $P(a_{2,3}=0)=0.50$.
If $a_{2,3}=0$, then for the sum of Row 2 to equal 3, the remaining 4 elements in Row 2 must sum to 3. The probability is:
$P(\text{Row 2 sum}=3\mid a_{2,3}=0)=\left(\genfrac{}{}{0ex}{}{4}{3}\right){0.5}^3{0.5}^1=4\times 0.0625=0.25$
Similarly, for the Column 3 sum to equal 3, the remaining 4 elements in Column 3 must sum to 3:
$P(\text{Column 3 sum}=3\mid a_{2,3}=0)=\left(\genfrac{}{}{0ex}{}{4}{3}\right){0.5}^3{0.5}^1=0.25$
The conditional joint probability for Case 2 is:
$P(R=3\cap C=3\mid a_{2,3}=0)=0.25\times 0.25=0.0625$

Total Probability Calculation:
Using the law of total probability, we combine the two cases:
$P(R=3\cap C=3)=P(R=3\cap C=3\mid a_{2,3}=1)P(a_{2,3}=1)+P(R=3\cap C=3\mid a_{2,3}=0)P(a_{2,3}=0)$
$P(R=3\cap C=3)=(0.140625\times 0.50)+(0.0625\times 0.50)$
$P(R=3\cap C=3)=0.0703125+0.03125=0.1015625$

Rounding this probability to two decimal places gives approximately 0.10.