A pair of dice is thrown until either a 4 or 6 appears.Find the probability that a 6 occurs first.

Post edited by: nivedita, at: 2008/03/23 23:16

the question is not clear....is it a pair of dice or a single one?
when u say either a 4 or a 6 appears, does that mean including both the dice or just any of the two dice?
from what i've understood in case there are 2 dice, there are three cases possible:

a) 4 appears first on any of the two dice
b) 6 appears first on any of the two dice
c) both 4 and 6 appear on the same throw.

lets first calculate the probability of case (c). At each throw of two dice, the probability of not getting either 4 or 6 is 2/3*2/3 = 4/9. The probability of getting both 4 and 6 is 2*1/6*1/6 = 1/18.
Therefore, prob. of getting 4 and 6 simultaneously on nth throw (will be a G.P. expression):
1/18 + 4/9*1/18 + (4/9)^2*1/18 + ...... = 1/10
(hope calculation is correct!)

Now, Prob of case (a) + Case (b) = 1 - 1/10 = 9/10
Therefore, prob of case (a) = prob of case (b) = 9/20

Therefore, ur answer shud be 9/20. IS it right?

when a pair of dice is thrown and we need 6 occurs first ,it actually means we need sum of the numbers on the top surface of two dices to be six before 4.

Post edited by: nivedita, at: 2008/03/24 11:11

That case is even more easy to solve.

There are 3 cases where sum can be 4, i.e. (1,3) (2,2) and (3,1) so probability of sum being 4 is 3/36 = 1/12

There are 5 cases were sum can be 6, i.e. (1,5) (2,4) (3,3) (4,2) and (5,1) so probability of sum being 6 is 5/36.

Therefore, probability of sum being any number other than 4
and 6 is 1 - (3/36 + 5/36) = 28/36 = 7/9

Now prob. of 6 occurring before 4 is:

5/36 + 7/9*5/36 + (7/9)^2*5/36 + ..... = 5/8

Is this answer correct?

the answer is correct.