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

Question

Aicharya · Accepted Answer

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

i n 1, 2, 3, 4, 6, 7, 8 o r 9 i s \frac{8}{10} . T h e r e f o r e t h e p r o b a b i l i t y t h a t t h e l a s t d i g i t o f t h e

product of

n

integer in

1, 2, 3, 4, 6, 7, 8 o r 9 i s {(\frac{4}{5})}^{n} .

The probability for an integer to

end in

1, 3, 7

or

9 i s \frac{4}{10} = \frac{2}{5}

Therefore the probability for the product of

n

positive integers to end in 1, 3, 7 or 9 is

{(\frac{2}{5})}^{n}

Hence the required probability

= {(\frac{4}{5})}^{n} - {(\frac{2}{5})}^{n} = \frac{4^{n} - 2^{n}}{5^{n}}

If n positive integers are taken at random and multiplied together, the probability that the last digit of…