Let P = begin{bmatrix} 1 & 2 & 3 4 & 5 & 6 | vedclass

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(B) The matrix $P$ has $9$ elements: $5$ odd $(1, 3, 5, 7, 9)$ and $4$ even $(2, 4, 6, 8)$.Total ways to select $3$ elements is $\binom{9}{3} = \frac{

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$\frac{17}{21}$

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(B) The matrix $P$ has $9$ elements: $5$ odd $(1, 3, 5, 7, 9)$ and $4$ even $(2, 4, 6, 8)$.Total ways to select $3$ elements is $\binom{9}{3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.Event $A$: Sum of $3$ elements is odd. This happens if we select ($3$ odd) or ($1$ odd, $2$ even).Ways for $A = \binom{5}{3} + \binom{5}{1} 	imes \binom{4}{2} = 10 + 5 	imes 6 = 10 + 30 = 40$.So, $P(A) = \frac{40}{84} = \frac{10}{21}$.Event $B$: Selecting $3$ elements in a row or column. There are $3$ rows and $3$ columns, so $6$ ways.$P(B) = \frac{6}{84} = \frac{1}{14}$.For $P(A|B)$, we need elements in a row or column whose sum is odd.Rows: $R_1(1,2,3) 	o 	ext{sum } 6 (	ext{even})$, $R_2(4,5,6) 	o 	ext{sum } 15 (	ext{odd})$, $R_3(7,8,9) 	o 	ext{sum } 24 (	ext{even})$.Columns: $C_1(1,4,7) 	o 	ext{sum } 12 (	ext{even})$, $C_2(2,5,8) 	o 	ext{sum } 15 (	ext{odd})$, $C_3(3,6,9) 	o 	ext{sum } 18 (	ext{even})$.There are $2$ such sets ($R_2$ and $C_2$).So, $P(A \cap B) = \frac{2}{84}$.$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{2/84}{6/84} = \frac{2}{6} = \frac{1}{3}$.$P(A) + P(A|B) = \frac{10}{21} + \frac{1}{3} = \frac{10+7}{21} = \frac{17}{21}$.

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