Discussion Forum



1)

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 xi 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


A) $\frac{9}{105}$

B) $\frac{10}{105}$

C) $\frac{11}{105}$

D) $\frac{7}{105}$

Answer:

Option C

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 

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

     = $\frac{11}{105}$