CBSE Class 12-science Answered
When a pair of dice is thrown,
the sample space is
{(1,1),(1,2)...(1,6)
(2,1),(2,2)..(2,6)
........(6,6)}
There are 36 entries in this set, so 36 sample points.
The entries favourable to getting a sum of 9 are (3,6),(4,5),(5,4),(6,3)
i.e. four entries
So prob of getting a total of nine =prob of winning=4/36=1/9
So prob of not winning =1-(1/9)=8/9
If A starts the game , then he may win it in the 1st throw, or 3rd throw, or 5th throw and so on..
prob of A winning is equal to
prob of winning in 1st throw+ prob of winning in 3rd throw+ prob of winning in 5th throw+...(or translates into addition)
=(1/9)+[(8/9)*(8/9)*(1/9)]+[(8/9)*(8/9)*(8/9)*(8/9)*(1/9)]+..
=(1/9){1+(8/9)*(8/9)+(8/9)*(8/9)*(8/9)*(8/9)+...}
The quantity inside the curly bracket is the sum of an infinite G.P. with first term as 1 and C.R. as 64/81
So the required answer is ( using the sum to infinity is equal to a/1-r)
=(1/9) {1/(1-64/81)
=(1/9) {81/17)
=9/17