If ‘n‘ integers taken at random and multiplied together, then what is the probability that the last digit of the product is 1, 3, 7 or 9 ?

• Ans: (2/5)^n
  Last digit may be 0,1,2,3,4,5,6,7,8,9 = 10
  condition is 1,3,7,9 = 4
  and n random no. so
  probability is 4^n/10^n=(4/10)^n=(2/5)^n
• 10 years agoHelpfull: Yes(22) No(1)
• probability will be 4/10 or 2/5
  as last digit can be any no between 0 to 9
  no of total cases will be 10
  prob=4/10
• 10 years agoHelpfull: Yes(9) No(1)
• as the last digit can be only when multipied with 1 only so the are only 4 no. out 10 so P is 2/5
• 10 years agoHelpfull: Yes(3) No(1)
• If the last digit in the product is to 1,3,7,9 the last digit in all n numbers should not be 0 and 5 and last digit of all numbers should not be selected exclusively from the set of numbers 2,4,6,8
  favourable no. of cases=(8^n)-(4^n)
  but generally last digit can be anyone of 0,1,2,3,4,5,6,7,8,9
  so,total no. of ways =10^n
  prob={(8^n)-(4^n)}/(10^n)
• 10 years agoHelpfull: Yes(1) No(1)
• 2n/5n
  There a 4 probable case of these numbers so 4n/10n and n be any random integer
  
• 10 years agoHelpfull: Yes(0) No(0)