GATE DA 2026 | Question: 54

Question

Let $A_{5 \times 5}$ be a matrix such that each of its elements follows $\operatorname{Bernoulli}(p=0.50)$ 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)

Comments

These questions are already posted why adding new ones instead of editing the old ones?
 

Answer 1

$\because$ Each cell follows bernoulli, hence the whole problem is Binomial(np)So,

Since the sum necessarily needs to be equal to 3 only we have two cases:

5 * 5 matrix

CASE-1: (orange cell is 1)

$$\Pr(\text{Case1}) = \Pr(\text{orange cell is 1})\times\Pr(\text{Sum is 3}\mid \text{orange cell is 1})$$

$$=\frac{1}{2} \times \underbrace{{}^4C_2 \times \left ( \frac{1}{2} \right )^2 \times \left ( \frac{1}{2} \right )^2}_{\text{for row}} \times \underbrace{{}^4C_2 \times \left ( \frac{1}{2} \right )^2 \times \left ( \frac{1}{2} \right )^2}_{\text{for column}}=\frac{9}{128}$$

CASE-2: (orange cell is 0)

$$\Pr(\text{Case 2}) = \Pr(\text{orange cell is 0})\times\Pr(\text{Sum is 3}\mid \text{orange cell is 0})$$

$$=\frac{1}{2} \times \underbrace{{}^4C_3 \times \left ( \frac{1}{2} \right )^3 \times \left ( \frac{1}{2} \right )^1}_{\text{for row}} \times \underbrace{{}^4C_3 \times \left ( \frac{1}{2} \right )^3 \times \left ( \frac{1}{2} \right )^1}_{\text{for column}}=\frac{1}{32}$$

Required probability: $\Pr(\text{Case 1})+\Pr(\text{Case 2}) = \dfrac{9}{128}+\dfrac{1}{32}=\boxed{\dfrac{13}{128} \approx 0.1016}$

 

 

 

---

Hence 0.1016 is accurate.

Answer 2

Row sum of the second row is:
$R_2$ = $A_{21} $+ $A_{22}$ + $A_{23}$ + $A_{24}$ + $A_{25}$
Column sum of the third column is:
$C_3 $= $A_{13}$ + $A_{23} $+ $A_{33}$ +$ A_{43}$+$A_{53}$

Element $A_{23}$ is shared between both the second row and the third column of the matrix.
$R_2$= $A_{23}$+$R'_2$          (where $R'_2$= $A_{21} $+ $A_{22}$ + $A_{24}$ + $A_{25}$)
$C_3$= $A_{23}$+$C'_3$           (where $C'_3$=$A_{13}$ + $A_{33} $ + $ A_{43}$ + $A_{53}$)

---

We want to find $P(R_2 = 3 \text{ and } C_3 = 3)$

Case 1: $A_{23}$=0
$R'_2$ = 3 & $C'_3$ = 3

Probability of getting a sum of 3 from 4 independent Bernoulli(0.5) trials is:
$P(X=3)=\binom{4}{3} (0.5)^3 (0.5)^1$
                    $= 4 \times (0.5)^4 = 4 \times 0.0625 = 0.25$

Since $R_2'$ ​ and $C_3' $ are independent So their joint probability is:
$P(R_2' = 3 \text{ and } C_3' = 3) = 0.25 \times 0.25 = 0.0625$

The probability of this case:
$P(A_{23} = 0) \times 0.0625 = 0.5 \times 0.0625 = 0.03125$

---

Case 2:$A_{23}$=1
$R'_2$ = 2 & $C'_3$ = 2
Probability of getting a sum of 2 from 4 independent Bernoulli(0.5) trials is:
$P(X=2)=\binom{4}{2} (0.5)^2 (0.5)^2$
                    $= 6 \times (0.5)^4 = 4 \times 0.0625 = 0.375$
Since $R_2'$ ​ and $C_3' $ are independent So their joint probability is:
$P(R_2' = 2 \text{ and } C_3' = 2) = 0.375 \times 0.375 = 0.140625$


The probability of this case:
$P(A_{23} = 1) \times 0.140625 = 0.5 \times 0.140625$ $= 0.0703125$

---

Adding Both Case Probability:
                                               $ = $  $0.03125+0.0703125$
                                                $=0.1015625$

Answer 3

Solution:

Each entry of the \(5 \times 5\) matrix is independent
Bernoulli\((p=\tfrac12)\).

We need:

Row 2 sum = 3  

Column 3 sum = 3  

 

Step 1: Identify overlap

 

Row 2 has 5 cells. Column 3 has 5 cells.  

They share one common cell \(A_{2,3}\).

Thus total distinct cells involved:

\[
5 + 5 - 1 = 9
\]

Let
\[
X = A_{2,3}.
\]

The only dependence between the row sum and column sum comes from this common cell.

 

Step 2: Condition on the overlap

Case 1: \(X = 1\)

\[
P(X=1) = \frac12
\]

If \(X=1\),

Row 2 must have 2 more ones among remaining 4 cells:

\[
\binom{4}{2}\left(\frac12\right)^4
=
\frac{6}{16}
=
\frac{3}{8}
\]

Column 3 must have 2 more ones among its remaining 4 cells:

\[
\frac{3}{8}
\]

These 8 cells are distinct and independent,
so joint probability:

\[
\frac{3}{8} \times \frac{3}{8}
=
\frac{9}{64}
\]

Multiply by \(P(X=1)\):

\[
\frac12 \times \frac{9}{64}
=
\frac{9}{128}
\]

 

Case 2: \(X = 0\)

\[
P(X=0) = \frac12
\]

Row 2 must have 3 ones among remaining 4 cells:

\[
\binom{4}{3}\left(\frac12\right)^4
=
\frac{4}{16}
=
\frac{1}{4}
\]

Column 3 must have 3 ones among its remaining 4 cells:

\[
\frac{1}{4}
\]

Joint probability:

\[
\frac{1}{4} \times \frac{1}{4}
=
\frac{1}{16}
\]

Multiply by \(P(X=0)\):

\[
\frac12 \times \frac{1}{16}
=
\frac{1}{32}
\]

Step 3: Add both cases

\[
\frac{9}{128} + \frac{1}{32}
=
\frac{9}{128} + \frac{4}{128}
=
\frac{13}{128}
\]

\[
\frac{13}{128} = 0.1015625
\]

Rounded to two decimal places:

\[
\boxed{0.10}
\]