Question

Let \( P = \begin{bmatrix} 1 & 2 & 3 \\4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \) be a matrix. Three elements of this matrix P are selected at random. A is the event of having the three elements whose sum is odd. B is the event of selecting the three elements which are in a row or column. Then \( P(A) + P(A|B) = \)

• A. 221/420
• B. 17/21
• C. 21/20
• D. 3/2

Hint:

For small sample spaces, listing outcomes is faster.

Answer 1

The Correct Option is B

Step 1: Calculation Process P(A):
Total selections \( \binom{9}{3} = 84 \). Odd sum requires 3 Odd or 2 Even + 1 Odd. Odds: \{1,3,5,7,9\} (5). Evens: \{2,4,6,8\} (4). Ways: \( \binom{5}{3} + \binom{4}{2}\binom{5}{1} = 10 + 30 = 40 \). \( P

(A) = 40/84 = 10/21 \).
Step 2: Calculation Process P(A|B):
B has 6 outcomes (3 rows, 3 cols). Outcomes with Odd sum: Row 2 (4+5+6=15), Col 2 (2+5+8=15). Total 2. \( P(A|B) = 2/6 = 1/3 \).
Step 3: Sum:
\( 10/21 + 1/3 = 10/21 + 7/21 = 17/21 \).
Step 3: Required Answer:
17/21.