A die with x sides has consecutive integers on its sides.If the probability of not getting a 4 on either of the 2 tosses is 36/49, how many sides does the die have? a. 4 b. 5 c. 7 d. 8 e. 13

ruplun wrote:A die with x sides has consecutive integers on its sides.If the probability of not getting a 4 on either of the 2 tosses is 36/49, how many sides does the die have? a. 4 b. 5 c. 7 d. 8 e. 13 IMO C I AM NOT SURE BUT explanation if suppose there are 7 sides then probability of getting 4 =1/7 probability of not getting 4=1-1/7 =6/7 probability pf not getting 4 for two times =6/7*6/7=36/49 thus die has 7 sides. if i am wrong plz correct me

let the die has n sides then probability of getting 4 = 1/n probability of not getting 4 = 1-(1/n) = (n-1)/n probability of not getting 4 twice = ((n-1)/n)*((n-1)/n) = sqr((n-1)/n) = 36/49 i.e. (n-1)/n = 6/7 on solving this n = 7 answer is c. 7

I just saw 7^2 in the denominator and chose C, knowing that in most of the probability-of-die-rolls questions I have faced, the solution often had 6^n in the denominator (n = the number of rolls of a standard dice).

Its quite simple. Let there be n sides of die. so probability of getting 4 in one throw = 1/n and probability of not getting 4 in one throw = (n-1)/n so probability of not getting a 4 i n both the 2 throw = {(n-1)/n}^2 --> {(n-1)/n}^2 = 36/49 --> {(n-1)/n} = 6/7 --> n = 7. Hence Ans C