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(C) The last digit of any number can be any of the $10$ digits: ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$.For the product of four numbers to have a last digit

Vedclass · Accepted Answer

$\frac{16}{625}$

Vedclass · Answer

(C) The last digit of any number can be any of the $10$ digits: ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$.For the product of four numbers to have a last digit of $1, 3, 5,$ or $7$,each of the four numbers must have a last digit that is odd and not $9$ (since $9 	imes 9 = 81$,which ends in $1$,but $9$ is not in the set ${1, 3, 5, 7}$).Actually,for the product to end in $1, 3, 5,$ or $7$,each individual number must end in one of these digits.There are $4$ favorable digits ${1, 3, 5, 7}$ out of $10$ possible digits for each number.The probability for one number is $P = \frac{4}{10} = \frac{2}{5}$.Since there are four independent numbers,the probability that all four have a last digit in ${1, 3, 5, 7}$ is $\left( \frac{2}{5} ight)^4 = \frac{16}{625}$.

If any four numbers are selected and they are multiplied,then the probability that the last digit of the product will be $1, 3, 5$ or $7$ is

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