Discussion Forum

Box I contains three cards bearing numbers 1,2,3: box II  contains five cards bearing numbers 1,2,3,4,5:  and box III  contained 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 numbers on the cars drawn from i th box i=1,2,3.

The probability that x₁,x₂,x₃  are in an arithmetic progression is

Answer:

Explanation:

 Plan   x₁,x₂,x₃    atr in A.P

 then x₂-x₁=x₃-x₂

  $\Rightarrow$     2 x₂= x₁+x₃

 $\therefore$   x₁,x₂,x₃  are in A.P

  x₁₊ x₃   =2x₂

  So, x₁ +x₃   should be even number

 Either both x₁ and x₃ are odd or both are even.

$\therefore$   Required probability 

       $=\frac{^{2}C_{1}\times^{4}C_{1}+^{1}C_{1}\times^{3}C_{1}}{^{3}C_{1}\times^{5}C_{1}\times^{7}C_{1}}$

     = $\frac{11}{105}$