IIT JEE , AIEEE Forum - Mathematics , Algebra

danny_007

a special dice is so constructed that the probabilities of throwing 1,2,3,4,5,6, are

(1-k)/6 , (1+2k)/6 , (1-k)/6 , ( 1 +k)/6 , (1-2k)/6 and (1+k)/6 resp

If 2 such dice are thrown and the probability of getting the sum = 9 lies between 1/9 and 2/9 , find the set of integral values of k

premsharma

Probability for sum 9 =

2*P(6)*P(3) + 2*P(5)*P(4) = (1-k^2)/18 + (1-k-k^2)/18

if this lies between 1/9 and 2/9.....then only integral value of k that seems possible is 0...
is this right?

danny_007

yes u r correct but more explanation needed please

computer001

ways of gettin 9:
3:6 6:3 5:4 4:5
total prob =
(1-k)(1+k)/36 +(1-k)(1+k)/36+ (1+k)(1-2k)/36 +(1+k)(1-2k)/36=
(2-3k^2-k)/18
so v get 2< 2-3k^2-k<4
1st:
k(1+3k)<0 so -1/3<k<0
2nd:
3k^2 + k +2>0 here a>0 D<0 hence always positive...
hence ans is:
-1/3<k<0

Ask & Discuss Questions with Community & Experts