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

• possible digits at the end of each no is 1,3,7,9 thus probablity for one is (4/10)
  we have 6 numbers so probablity will be (4/10)^6
• 10 years agoHelpfull: Yes(35) No(7)
• product should be odd...if multiplied by 5 we will get 5 or 0 at end..so let us exclude 5...then remaining numbers= even(2,4,6,8),odd=(1,3,7,9)
  odd*odd=odd
  even*even=even
  odd*even=even
  => all numbers should be odd=number of chance=4^6
  all possible chances=9^6
  probability=(4/9)^6
  
• 10 years agoHelpfull: Yes(24) No(8)
• can anyone explain in detail
• 10 years agoHelpfull: Yes(8) No(2)
• @NIKHIL PANDE WHAT IS THE EXACT ANS ........DO U KNOW???
• 10 years agoHelpfull: Yes(7) No(1)
• product should not end with 5 so the digits possible are 1,3,7,9.
  Inorder to satisfy this condition the product should be divisible by 2 and 5.
  Hence the probability of a no.to be divisible by 2 & 5 is 6/10(0,2,4,5,6,8).
  Hence the prob for ending with 1,3,7,9 is 1-(6/10)=4/10=2/5.
  For 6 digits the prob is (2/5)^6=64/15625
• 10 years agoHelpfull: Yes(6) No(14)
• Let numbers are 1,2,3,4,5,6....
  out of all cases only (1*3)and (3*3)end with odd digit other then 5
  so ans is 1/3
• 10 years agoHelpfull: Yes(1) No(5)
• possible digits at the end of multiplication is 1,3,7,9
  total(1,2,3,4,5,6,7,8,9,0)
  probability that products end with odd other than 5 is (4/10)
• 10 years agoHelpfull: Yes(1) No(3)
• in any number the last digit can be 0,1,2,3,4,5,6,7,8,9 .We want the last digit in the product is an odd digit other than 5 i.e; it is any one of the digits 1,3,7,9. This means that the product is not divisible by 2 or 5.The probability that a number is divisible by 2 or 5 is 6/10.
  [since , in that case the last digit can be one of 0,2,4,5,6,8]
  therefore, the probability that the number is not divisible by 2 or 5 = 1- 6/10 = 4/10 = 2/5
  in order that the product is not divisible by 2 or 5 none of the constituent numbers should be divisible by 2 or 5 and its probability = (2/5)^6 = 64/390625
  
• 9 years agoHelpfull: Yes(1) No(1)
• the event will occurs when all the numbers selected are ending in 1, 3, 7 or 9.
  if we take numbers between 1 to 10(both inclusive), we will have a positive occurrence if each of the six numbers selected are either 1, 3, 7 or 9....
  
  the probability of any number selected being either of these 4 = 4/10 (4 possible events out of 10 possibilities)
  answer is ( 0.4 )^6....
• 9 years agoHelpfull: Yes(1) No(1)
• only 0dd*odd gives odd
  so
  all 6 no. need to be odd
  nd also to eliminate no ending with 5 ..
  we cant use no. ednding in 5 as any odd*5=end digit 5;
  so odd digits that can be taken r 1,3,7,9
  nd tatal digits 10
  so ans 4/10
  
  
• 10 years agoHelpfull: Yes(0) No(4)
• total no. of cases=total no. of odd digits that might be the last digit (1,3,5,7,9)=5
  total no. of favorable cases=odd digits excluding 5=(1,3,7,9)=4;
  so,probability=4/5;
  pls provide optiions;@ nikhil pande
• 9 years agoHelpfull: Yes(0) No(1)
• Out of ten positive integers, there are 5 odd and 5 even integers. Odd integers other than 5 are 1,3,7,9.
  So, probability of selecting any odd integer other than 5 is given as : 4/10
  As the selected integer can repeat itself, the answer is (4/10)^6
• 7 years agoHelpfull: Yes(0) No(1)
• The event will occurs when all the numbers selected are ending in 1, 3, 7 or 9.
  If we take numbers between 1 to 10 (both inclusive), we will have a positive occurrence if each of the six
  numbers selected are either 1, 3, 7 or 9.
  The probability of any number selected being either of these 4 is 4/10 (4 positive events out of 10
  possibilities)
  [Note: If we try to take numbers between 1 to 20, we will have a probability of 8/20 = 4/10. Hence, we
  can extrapolate up to infinity and say that the probability of any number selected ending in 1, 3, 7 or 9 so
  as to fulfill the requirement is 4/10.]
  Hence, answer = (0.4)^6
• 5 years agoHelpfull: Yes(0) No(0)