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Question

If n positive integers are taken at random and multiplied together, and pn is the probability that the last digit of the product is 2,4,6 or 8. Find 125×p3

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Solution

The last digit of the product will be 1,2,3,4,6,7,8 or 9 if and only if each of the n positive integers ends in any of these digits.
Now, the probability of an integer ending in 1,2,3,4,6,7,8 or 9 is 8/10=4/5.
Therefore, the probability that the last digit of the product of n integers is 1,2,3,4,6,7,8 or 9 is (4/5)n.
Next, the last digit of the product will be 1,3,7 or 9 if and only if each of the n positive integers ends in 1,3,7 or 9.
The probability for an integer to end in 1,3,7 or 9 is 4/10=2/5.
Therefore, the probability for the product of n positive integers to end in 1,3,7 or 9 is (2/5)n.
Hence, the probability of the required event is
(45)n(25)n=4n2n5n
For n=3
P3=432353=648125=56125125×P3=56

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