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 $x_1,x_2$ and $x_3$ are in 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

$\frac{11}{105}$

Since, $x_1,x_2,x_3$ are in AP.
$\therefore x_1+x_3=2x_2$
So,$x_1+x_3$ should be even number.
Either both $x_{1}and{x}_3$ are odd or both are even.
$\therefore Requiredprobability=\frac{{}^2{C}_1{\times }^4{C}_1{+}^1{C}_1{\times }^3{C}_1}{{}^3{C}_1{\times }^5{C}_1{\times }^7{C}_1}=\frac{11}{105}$