Question

Box $1$ contains three cards bearing numbers $1,2,3;$ box $2$ contains five cards bearing numbers $1,2,3,4,5;$ and box $3$ contains seven cards bearing numbers $1,2,3,4,5,6,7.$ A card is drawn from each of the boxes. Let $x_i$ be the number on the card drawn from the $i^{th}$ box, $i=1,2,3.$

The probability that $x_1+x_2+x_3$ is odd, is

• A. $\frac{29}{105}$
• B. $\frac{53}{105}$
• C. $\frac{57}{105}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{53}{105}$

Sum of $x_1+x_2+x_3$ to be odd,
Case 1:
When all is odd
Probability $=\frac{2\times 3\times 4}{3\times 5\times 7}=\frac{8}{35}$

Case 2:
When one of them is odd,
When $x_1$ is odd
Probability $=\frac{2\times 2\times 3}{3\times 5\times 7}=\frac{4}{35}$
When $x_2$ is odd
Probability $=\frac{1\times 3\times 3}{3\times 5\times 7}=\frac{3}{35}$
When $x_3$ is odd
Probability $=\frac{1\times 2\times 4}{3\times 5\times 7}=\frac{8}{105}$
Therefore the required probability will be
$=\frac{8}{35}+\frac{4}{35}+\frac{3}{35}+\frac{8}{105}=\frac{53}{105}$