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

• 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.
  
• 9 years agoHelpfull: Yes(17) No(4)
• 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%
• 7 years agoHelpfull: Yes(17) No(5)
• P(5)= 4/36
  P(7)= 6/36
  P(another 5 before 7)= 1-(10/36)
  The answer is 26/36
• 10 years agoHelpfull: Yes(8) No(24)
• 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%
• 6 years agoHelpfull: Yes(1) No(1)
• your answer is unique yoga priya....
• 6 years agoHelpfull: Yes(0) No(4)
• my ans is 40%
• 4 years agoHelpfull: Yes(0) No(0)