Let S be the sample space of all 3 times 3 ma | vedclass

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(C) The total number of entries in a $3 	imes 3$ matrix is $9$. Since the sum of entries is $7$ and entries are from $\{0, 1\}$,there must be exactly

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$0.50$

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(C) The total number of entries in a $3 	imes 3$ matrix is $9$. Since the sum of entries is $7$ and entries are from $\{0, 1\}$,there must be exactly $7$ ones and $2$ zeros.The number of ways to choose positions for the $2$ zeros is $n(E_2) = \binom{9}{2} = \frac{9 	imes 8}{2} = 36$.For $\operatorname{det} A = 0$,the matrix must have at least one row or column consisting entirely of zeros,or be linearly dependent. Since there are only $2$ zeros,the determinant is $0$ if and only if the two zeros are in the same row or the same column.Number of ways to place $2$ zeros in the same row: There are $3$ rows,and in each row,there are $\binom{3}{2} = 3$ ways to place the zeros. So,$3 	imes 3 = 9$ ways.Number of ways to place $2$ zeros in the same column: There are $3$ columns,and in each column,there are $\binom{3}{2} = 3$ ways to place the zeros. So,$3 	imes 3 = 9$ ways.Thus,$n(E_1 \cap E_2) = 9 + 9 = 18$.The conditional probability is $P(E_1 \mid E_2) = \frac{n(E_1 \cap E_2)}{n(E_2)} = \frac{18}{36} = 0.50$.

List-$I$	List-$II$
$A$. $P(\frac{A}{B})$	$(i)$. $\frac{2}{15}$
$B$. $P(\bar{B})$	$(ii)$. $\frac{4}{15}$
$C$. $P(A \cap \bar{B})$	$(iii)$. $\frac{8}{15}$
$D$. $P(B \cap \bar{A})$	$(iv)$. $\frac{2}{3}$
	$(v)$. $\frac{3}{7}$

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