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USE CODE: EXAM25

				   

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

3 years ago

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Answers : (3)

										

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).

3 years ago
										

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


(1,1,1) (2,2,2)........(6,6,6).

3 years ago
										

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

3 years ago

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