GATE 2026 DA Q.64: solution and answer

GATE 2026 · DAQ.64 · 2 marks · NAT

Let $A_{5 \times 5}$ be a matrix such that each of its elements follows $B er n o u l l i (p = 0.50)$ 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)

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Answer and solution

ANSWER

The probability is $\frac{26}{256} \approx 0.10$ .

Official answer: 0.08 to 0.12Source: Official answer key, GATE 2026 DA (IIT Guwahati)

The second row and the third column of the $5 \times 5$ matrix share exactly one entry, $a_{23}$ . The other $4$ entries of the row and the other $4$ entries of the column are $8$ distinct entries, and all of them are independent $Bernoulli (\frac{1}{2})$ .
Condition on $a_{23} = x$ . Given $x$ , the row needs its other 4 entries to sum to $3 - x$ , and the column needs its other 4 entries to sum to $3 - x$ . These two events involve disjoint sets of entries, so they are independent given $x$ .
Let $S \sim Binomial (4, \frac{1}{2})$ be the sum of 4 such entries. Then $P (S = k) = (k 4) /16$ .
Case $x = 1$ , which has probability $\frac{1}{2}$ : both the row and the column need $S = 2$ . $P (S = 2) = \frac{6}{16}$ , so the contribution is $\frac{1}{2} (\frac{6}{16})^{2} = \frac{18}{256}$ .
Case $x = 0$ , which has probability $\frac{1}{2}$ : both the row and the column need $S = 3$ . $P (S = 3) = \frac{4}{16}$ , so the contribution is $\frac{1}{2} (\frac{4}{16})^{2} = \frac{8}{256}$ .
By total probability, $P = \frac{18 + 8}{256} = \frac{26}{256} \approx 0.1016$ .
Rounded to two decimal places, the answer is $0.10$ . (Treating the two events as independent would give $(\frac{10}{32})^{2} \approx 0.098 \approx 0.10$ here by coincidence, but that method is not valid because the events share $a_{23}$ .)

KEY IDEA

The row and the column overlap in one entry, $a_{23}$ . Condition on that entry. Given it, the other 8 entries are independent, so the row event and the column event are independent and each is a Binomial $(4, \frac{1}{2})$ probability.

WATCH OUT FOR

Multiplying $P (row sum = 3) \cdot P (column sum = 3) = (10/32)^{2}$ as if the events were independent. They are not, because they share $a_{23}$ . The result happens to round to the same $0.10$ here, but the method is wrong.

From the GATE 2026 DA paper. Every question of this paper · Download the full paper · All 2 Discrete distributions questions, with notes