A) \[\frac{{{4}^{n}}+{{2}^{n}}}{{{5}^{n}}}\]
B) \[\frac{{{4}^{n}}\times {{2}^{n}}}{{{5}^{n}}}\]
C) \[\frac{{{4}^{n}}-{{2}^{n}}}{{{5}^{n}}}\]
D) None of these
Correct Answer: C
Solution :
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 \[\frac{8}{10}.\] Therefore the probability that the last digit of the product of \[n\] integers in \[1,\,2,\,3,\,4,\,6,\,7,\,8\] or 9 is \[{{\left( \frac{4}{5} \right)}^{n}}.\] The probability for an integer to end in \[1,\,3,\,7\] or 9 is \[\frac{4}{10}=\frac{2}{5}.\] Therefore the probability for the product of \[n\] positive integers to end in \[1,\,3,\,7\] or 9 is \[{{\left( \frac{2}{5} \right)}^{n}}.\] Hence the required probability\[={{\left( \frac{4}{5} \right)}^{n}}-{{\left( \frac{2}{5} \right)}^{n}}=\frac{{{4}^{n}}-{{2}^{n}}}{{{5}^{n}}}\].You need to login to perform this action.
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