Let ‘S’ be the sample space of all 3×3 matrices with entries from set {0, 2}. Let the events. E1, E2 &E3 be given byE1={A∈S:|A|=0} E2={A∈S: Sum of entries of A is 14} E3={A∈S: Trace of A is 4}If a matrix is chosen at random from S, then PE1E2−PE3E2 is

n(E2)= arrangement of ‘7’ ‘2’ s and ‘2’ ‘0’ s = 9C2=9×82=36n(E1∩E2)= both zeros should be in a row or column.222222202→ In each row we have 3 waysIn total we have 18 ways⇒P(E1E2)=12 ⇒P(E3E2)=P(E3∩E2)P(E2) n(E2)=36 =3C1× 6C1=18⇒P(E3E2)=1836=12

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Let ‘S’ be the sample space of all $3 \times 3$ matrices with entries from set ${0, 2}$ . Let the events. $E_{1}, E_{2} & E_{3}$ be given by
$E_{1} = {A \in S : | A | = 0} E_{2} = {A \in S : S u m o f e n t r i e s o f A i s 14} E_{3} = {A \in S : T r a c e o f A i s 4}$
If a matrix is chosen at random from S, then $P (\frac{E_{1}}{E_{2}}) - P (\frac{E_{3}}{E_{2}})$ is

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Detailed Solution

$n (E_{2}) =$ arrangement of ‘7’ ‘2’ s and ‘2’ ‘0’ s = $^{9} C_{2} = \frac{9 \times 8}{2} = 36$
$n (E_{1} \cap E_{2}) =$ both zeros should be in a row or column.
$(\begin{matrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 0 & 2 \end{matrix}) \to$ In each row we have 3 ways
In total we have 18 ways