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 $x_1,x_2,x_3$ are in an arithmetic progression, is

• A. $\frac{9}{105}$
• B. $\frac{10}{105}$
• C. $\frac{11}{105}$
• D. $\frac{7}{105}$

Answer 1

The correct option is C $\frac{11}{105}$

$x_1,x_2,x_3$ are in $A.P.$
$\Rightarrow x_1+x_3=2x_2$
$\Rightarrow x_1+x_3=\text{even}$
$\therefore x_1,x_3$ both should be either even or odd
So, total number of favourable cases is ${}^1{c}_1.{}^3{c}_1+{}^2{c}_1.{}^4{c}_1=11.$
Required probability $=\frac{11}{3\times 5\times 7}=\frac{11}{105}$