A) \[\frac{1}{4}\]
B) \[\frac{1}{7}\]
C) \[\frac{1}{8}\]
D) \[\frac{1}{49}\]
Correct Answer: A
Solution :
Since \[m\] and \[n\] are selected between 1 and 100, hence sample space \[=100\times 100.\] Also \[{{7}^{1}}=7\], \[{{7}^{2}}=49\], \[{{7}^{3}}=343\], \[{{7}^{4}}=2401\], \[{{7}^{5}}=16807\] etc. Hence \[1,\,\,3,\,\,7\] and 9 will be the last digits in the powers of 7. Hence for favourable cases \[\begin{align} & n\,\,m\,\,\to \\ & \downarrow \\ \end{align}\] \[\begin{matrix} 1,\,\,1 & \,\,\,\,\,\,\,1,\,\,2 & \,\,\,1,\,\,3\,.......\,\,\,\,\,\,\,1, & 100 \\ 2,\,\,1 & \,\,\,\,\,\,2,\,\,2 & \,\,2,\,\,3\,.......\,\,\,\,\,\,\,2, & 100 \\ \end{matrix}\] ?????????????? \[\begin{matrix} 100,\,\,1 & 100,\,\,2 & 100,\,\,3\,.......\,100, & 100 \\ \end{matrix}\] For \[m=1;\,\,\,n=3,\,\,7,\,\,11.....97\] \[\therefore \] Favourable cases = 25 For \[m=2;\,\,\,n=4,\,\,8,\,\,12.....100\] \[\therefore \] Favourable cases = 25 Similarly for every \[m,\] favourable \[n\] are 25. \[\therefore \] Total favourable cases \[=100\times 25\] Hence required probability \[=\frac{100\times 25}{100\times 100}=\frac{1}{4}.\]
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