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(#M40021918) AIEEE Numerical Ability question PROBABILITY Keep an EYE

On a toss of two dice, A throws a total of 5. Then the probability that he will throw another 5 before he throws 7 is

Asked In AIEEE (11 years ago)

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Read Solution (6) :

siva rayalu  (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.

Is this solution Helpfull? Yes (17) | No (4)  

rathinamurugan  (8 years ago)

ans: 40%
Explanation:
total probabilities for getting 5 = 4/36
total probabilities for getting 7 = 6/36
Total Probability = 10/36
We need only 5, hence prob of getting only 5 is (4/36)/(10/36)
=40%

Is this solution Helpfull? Yes (17) | No (5)  

Yoga Priya  (11 years ago)

P(5)= 4/36
P(7)= 6/36
P(another 5 before 7)= 1-(10/36)
The answer is 26/36

Is this solution Helpfull? Yes (8) | No (24)  

aartee  (7 years ago)

total probabilities for getting 5 = 4/36
total probabilities for getting 7 = 6/36
Total Probability = 10/36
We need only 5, hence prob of getting only 5 is (4/36)/(10/36)
=40%

Is this solution Helpfull? Yes (1) | No (1)  

Deepak  (7 years ago)

your answer is unique yoga priya....

Is this solution Helpfull? Yes | No (4)  

lethaswi  (5 years ago)

my ans is 40%

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