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ans: 40%

**Alternative way - (Difficult)**

5 can be thrown in 4 ways (1,4), (2,3), (3,2), (4,1)

Hence the probability of throwing a 5 in a single throw with a pair of dice is 4/36 = 1/9

Similarly, 7 in 6 ways in a single throw with a pair of dice.

Hence the number of ways of throwing neither 5 nor 7 is 36 - ( 4 + 6 ) = 26

And the probability of throwing neither 5 nor 7 is ( 26/36 ) = ( 13/18 )

Hence the required probability

= ( 1/9 ) + ( 13/18 ) ( 1/9 ) + ( 13/18 )² ( 1/9 ) + ( 13/18 )³ ( 1/9 ) + .....

= ( 1/9 ) / [ 1 - ( 13/18 ) ] = 2/5 (using AP, GP formula)

EXPLANATION :

Since he has already thrown a 5, ( ie a number different from 7), he may throw 5 at the next attempt. The probability for which is ( 1/9 ) or he may throw 5 at the second attempt when he fails to throw either 5 or 7 at the first attempt, the probability for which is ( 13/18 ) ( 1/9 ) or at the third attempt the probability for which is ( 13/18 )² ( 1/9 ) and so on ........ Adding all the probabilities we get the answer.

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