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# A and B throw a pair of die.......

Asked by 19th March 2009, 9:43 AM

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

Answered by Expert 19th March 2009, 10:18 PM
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