Question

If four digits from $\{1,2,3,.....9\}$ are taken at random and multiplied together, then the chance that the last digit in the product be $1,3,7$ or $9$ is

• A. $\frac{9}{{}^9{P}_4}$
• B. $\frac{1}{10}$
• C. $\frac{1}{{}^9{C}_4}$
• D. ${\left(\frac{4}{9}\right)}^4$

Answer 1

The correct option is D ${\left(\frac{4}{9}\right)}^4$

$1,2,3,4,5,6,7,8,9$
For the product to end with $1,3,7,9$
none of the no. among the four should be an even no. so $2,4,6,8$ excluded also, multiple of 5 either end in 5 or 0. hence 5 also is excluded.
Thus any no. selected from $1,3,7,9$ and multiplied will end in $1,3,7,9$
$P\left(1or3or7or9\right)=\frac{4}{9}$
Hence answer is $(\frac{4}{9}{)}^4$