Question

Box I contain three cards bearing numbers 1,2,3; box II contains five cards bearing numbers 1,2,3,4,5; and box III 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

$\frac{53}{105}$

$Probability=\frac{Numberoffavourableoutcomes}{Numberoftotaloutcomes}$
As, $x_1+x_2+x_3$ is odd.
So, all may be odd or one of them is odd and other two are even.
$\therefore$ Required probability
$=\frac{{}^2{C}_1{\times }^3{C}_1{\times }^4{C}_1+{}^2{C}_1{\times }^2{C}_1{\times }^3{C}_1+{}^1{C}_1{\times }^3{C}_1{\times }^3{C}_1+{}^1{C}_1{\times }^2{C}_1{\times }^4{C}_1}{{}^3{C}_1{\times }^5{C}_1{\times }^7{C}_1}=\frac{24+12+9+8}{105}=\frac{53}{105}$