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if 4 whole numbers taken at random are multiplied together then the chance that the last digit in the product is 1,3,7,9 is
Asked by | 04 Feb, 2011, 08:40: AM
There are 10 digits 0,1,2,........9 any of which can occur in any number at the last place.
It is obvious that if the last digit in any of the four numbers is 0,2,4,5,6,8 then product of any such four numbers will not give a number having it's last digit as 1,3,7,9. Hence, it is necessary that the last digit in each of the four numbers must be any of the four digits 1,3,7,9.
Thus, for each of the four numbers,
the number of ways for the last digit = 10
favourable number of ways = 4
therefore, the probability that the last digit is any of the four numbers 1,3,7,9 = 4/10 = 2/5
Hence, the required probability that the last digit in each of the four numbers is 1,3,7,9 so that the last digit in their product is 1,3,7,9 = (2/5)4 = 16/625.
Answered by | 07 Feb, 2011, 10:08: AM

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