Latest TCS Aptitude Question SOLUTION: on a toss of two dice A throws a total of 5. what is the probability that he throws another 5 before he throws 7

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on a toss of two dice A throws a total of 5. what is the probability that he throws another 5 before he throws 7

Asked In TCS karthika (10 years ago)

Unsolved Read Solution (4)

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Read Solution (4) :

Tanvi  (10 years ago)

5 can be thrown in 4 ways and 7 in 6 ways in a single throw with a pair of dice. Hence number of ways of throwing neither 5 nor 7 is 36 - ( 4 + 6 ) = 26

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

And probabilty 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

EXPLAINATION :

Since he has already thrown a 5, ( ie a number different from 7 ) , he may throw 5 at the next attempt. The probabilty 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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manoj chauhan  (10 years ago)

I think it could be 1/9
because probability to getting a sum of 5 is (1,4),(4,1),(3,2),(2,3)
S=36
now p=4/36=>1/9

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sandya  (10 years ago)

ans 4/36
total exp r 36

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keerthana  (5 years ago)

can any1 explain me clearly.

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