1)

Three boys and two girls stand in a queue. The probability  that the number of boys ahead of every girl is at least  one more than the number of girls ahead  of her, is 


A) 1/2

B) 1/3

C) 2/3

D) 3/4

Answer:

Option A

Explanation:

Plan (i)  The number of arrangement of m distinct objects is given by n!.

 (ii) Probability =

                      NUmber of favourable outcomes/ Number of total out comes

  Total number of ways  to arrange 3 boys and 2 girls  are 5!

 According to given condition , following cases may be

   BGGBB

   GGBBB

   GBGBB

   GBBGB

   BGBGB

  So, number of favourable ways

      = 5! x 3! x 2!

  = 60

 $\therefore$   Required probability

                    =    $\frac{60}{120}=\frac{1}{2}$