6 positive number are taken at random and multiplied together. Then what is the probability that the product ends in an odd digit other than 5

• Total no of ways is 10
  Reqired selections
  (i.e the selected digit must end with 1,3,7 or 9)
  = 4
  hence probability= 4/10= 2/5
• 11 years agoHelpfull: Yes(12) No(4)
• number ending with 1,3,7,9 should be taken
  so (2/5)^4=16/625
• 11 years agoHelpfull: Yes(5) No(15)
• Answer is (4/10)^6 ....
  Total Ways of getting any positive number = 10^6
  To get an odd number (not having 5 in unit's place) we'll need - 1,3,7,9 and each of which will appear in 4^6 ways in unit's place.
  
  so answer - (4/10)^6
• 11 years agoHelpfull: Yes(5) No(0)
• can any one tell me solution
• 11 years agoHelpfull: Yes(1) No(11)
• Ans is ---> 64/15625
• 10 years agoHelpfull: Yes(1) No(0)
• shouvik ghosh your answer is correct..i checked it with another website
  16/625 but i didnt understand how you did it. could you please explain?
• 9 years agoHelpfull: Yes(0) No(1)
• the unit digit must be odd according to the qs.....
  so we hv 1,3,7,9...this 4 odd digit out of 10 positive digits excluding 5....
  so probability of avalibilty of odd digits=4/10=2/5
  but this is not the ans...bcz we cannot put the 4 digits at a time in unit digits place...its impossible...
  so there any one out of the 4 digits can be...
  so the probabilty is the no of ways in which that is possible and that is=(2/5)^4=16/625
• 9 years agoHelpfull: Yes(0) No(0)