Question

If $n$ positive integers are taken at random and multiplied together, and $p_n$ is the probability that the last digit of the product is $2,4,6$ or $8$. Find $125\times p_3$

Answer 1

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
${\left(\frac{4}{5}\right)}^n-{\left(\frac{2}{5}\right)}^n=\frac{4^n-2^n}{5^n}$
For $n=3$
$P_3=\frac{4^3-2^3}{5^3}=\frac{64-8}{125}=\frac{56}{125}\Rightarrow 125\times P_3=56$