given that n sided fair has sides labelled 1 through n where n>3 ;
If the die is rolled once, is the probability of rolling a "8" greater than 2/31
die has 6 sides in this case side would be ( 4,5,6,7,8,9) ; (5,6,7,8,9,10) ; ( 6,7,8,9,10,11) ; so on till ( 8,9,10,11,12,13)
#1
If the die is rolled twice, the probability that the numbers obtained are different is greater than 5/6.
it means ; 1-(1/n)^2 >5/6 ;
6n^2-6>5n^2
n^2>6
n can be any value > 3 ;
we see that getting value of 1/8 for given set of die from 1/9 to 1/13 ; .11 to .076 which in all cases is > 2/31;.064
sufficient
#2
If the die is rolled twice, the probability that the numbers obtained are equal is greater than 1/8.
(1/n)^2 >1/8
this is only possible when value of n = 1 or 2
but given that n>3 ;
it means that answer can be yes and no for values of n where 8 is included
insufficient
option A is correct Bunuel wrote:
An n-sided fair die has sides labeled with the numbers 1 through n, inclusive, where n > 3. If the die is rolled once, is the probability of rolling a "8" greater than 2/31 ?
(1) If the die is rolled twice, the probability that the numbers obtained are different is greater than 5/6.
(1) If the die is rolled twice, the probability that the numbers obtained are equal is greater than 1/8.