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If the four positive integers are selected randomly from the set of positive integers, then the probability that the number $1,3,7$ and 9 are in the unit place in the product of 4 -digit, so selected is
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The correct answer is:
$\frac{16}{625}$
The number of digits on unit place of any number $=10$
$\therefore$
$$
n(S)=10
$$
The necessary condition for becoming the digits $1,3,5$ or 7 at the unit place of product of four numbers that the digits $1,3,5$ or 7 at unit place of every number.
$$
\begin{array}{ll}
\therefore & n(A)=4 \\
\therefore & P(A)=\frac{4}{10}=\frac{2}{5}
\end{array}
$$
So, required $y=\left(\frac{2}{5}\right)^{4}=\frac{16}{625}$
$\therefore$
$$
n(S)=10
$$
The necessary condition for becoming the digits $1,3,5$ or 7 at the unit place of product of four numbers that the digits $1,3,5$ or 7 at unit place of every number.
$$
\begin{array}{ll}
\therefore & n(A)=4 \\
\therefore & P(A)=\frac{4}{10}=\frac{2}{5}
\end{array}
$$
So, required $y=\left(\frac{2}{5}\right)^{4}=\frac{16}{625}$
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