Question

A$_{5\times5$ 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-}

Hint:

When calculating the joint probability of events that are not independent, identify the source of dependence (here, the shared matrix element). Conditioning on the possible states of this shared part is a powerful technique to break the problem down into independent sub-problems.

Answer 1

Correct Answer: 0.1

Step 1: Understanding the Question:
We have a 5x5 matrix where each of the 25 entries is independently chosen to be 0 or 1 with equal probability (P=0.5). We need to find the probability that two events occur simultaneously: 1. The sum of the 5 elements in the second row is 3. 2. The sum of the 5 elements in the third column is 3.
Step 2: Key Formula or Approach:
The two events are not independent because they share the element at the intersection, A(2,3). We can solve this by conditioning on the value of this common element. Let A be the event that row 2 sum is 3, and B be the event that column 3 sum is 3. P(A $\cap$ B) = P(A $\cap$ B | A(2,3)=1)P(A(2,3)=1) + P(A $\cap$ B | A(2,3)=0)P(A(2,3)=0).
Step 3: Detailed Explanation:
Let X = A(2,3). P(X=1) = 0.5 and P(X=0) = 0.5.
The number of 1s in a set of 'n' independent Bernoulli(0.5) trials follows a Binomial distribution B(n, 0.5).
The probability of getting 'k' ones is $C(n,k)(0.5)^n$.
Case 1: X = 1 (A(2,3) = 1)
- For the sum of row 2 to be 3, the other 4 elements in that row must sum to 2.
The probability is $P(\text{sum of 4 elements}=2) = C(4,2)(0.5)^4 = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8}$.
- For the sum of column 3 to be 3, the other 4 elements in that column must also sum to 2. The probability is $P(\text{sum of 4 elements}=2) = C(4,2)(0.5)^4 = \frac{6}{16} = \frac{3}{8}$.
- Since these two sets of 4 elements are disjoint, their probabilities are independent. The contribution from this case is: \[ P(\text{Case 1}) = P(X=1) \times \left(\frac{3}{8}\right) \times \left(\frac{3}{8}\right) = \frac{1}{2} \times \frac{9}{64} = \frac{9}{128} \] Case 2: X = 0 (A(2,3) = 0) - For the sum of row 2 to be 3, the other 4 elements in that row must sum to 3.
The probability is $P(\text{sum of 4 elements}=3) = C(4,3)(0.5)^4 = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$.
- For the sum of column 3 to be 3, the other 4 elements in that column must also sum to 3. The probability is $P(\text{sum of 4 elements}=3) = C(4,3)(0.5)^4 = \frac{4}{16} = \frac{1}{4}$.
- The contribution from this case is:
\[ P(\text{Case 2}) = P(X=0) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{2} \times \frac{1}{16} = \frac{1}{32} \]
Total Probability:
The total probability is the sum of the probabilities from the two mutually exclusive cases.
\[ P(\text{Total}) = P(\text{Case 1}) + P(\text{Case 2}) = \frac{9}{128} + \frac{1}{32} = \frac{9}{128} + \frac{4}{128} = \frac{13}{128} \]
Step 4: Final Answer:
The probability is $\frac{13}{128}$. Converting this to a decimal gives $13 \div 128 = 0.1015625$. The question in the PDF gives the answer as 0.10.