The no.s (a,b,c) are selected by throwing a dice thrice.then the probability that (a,b,c) are in A.P. is?

i am geting the ans to be 18/216 but the ans is 21/216

Hi Debadutta,

(a,b,c) can be

(1,2,3) -------- 2 possibilities (3,2,1) is the other order

(2,3,4) -------- 2 possibilities

.....     --------- -do-

.....

(4,5,6) -------- -do-

it could be

(1,3,5) -------- 2 possibilities

(2,4,6) -------- 2 possibilities

or it could be

(1,1,1)

(2,2,2)

....

....

(6,6,6).

So total number of ways = 2*6 + 6 = 18.

Your answer is very fine.

Have confidence in your working. It helps for competitive exam prearation.

21/216 is not the right answer.

Regards,

Ashwin (IIT Madras).

How can This be an A.P.Minimum C.D for an AP is 1.

(1,1,1) (2,2,2)........(6,6,6).

the answar is  18/216, there cant be more than 18 combos of AP for 3 die