Question

Two numbers ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ are selected successively without replacement in that order from the integers $1$ to $10$. The probability that $\frac{a}{b}$ is an integer, is

• A. $\frac{17}{45}$
• B. $\frac{1}{5}$
• C. $\frac{17}{90}$
• D. $\frac{8}{45}$

Answer 1

Answer: (C)

$\frac{17}{90}$

Given,two numbers are selected without replacement from integers $1$ to $10$.

Now, favourable cases for which $\frac{a}{b}$ will be an integer are:

$\frac{2}{1}=2$;$\frac{3}{1}=3$;$\frac{4}{1}=4$;$\frac{4}{2}=2$;$\frac{5}{1}=5$;

$\frac{6}{1}=6$;$\frac{6}{2}=3$;$\frac{6}{3}=2$;$\frac{7}{1}=7$;$\frac{8}{1}=8$;

$\frac{8}{2}=4$;$\frac{8}{4}=2$;$\frac{9}{1}=9$;$\frac{9}{3}=3$;$\frac{10}{1}=10$;

$\frac{10}{2}=5$;$\frac{10}{5}=2$

$\therefore$, favourable number of cases$=17$

Total cases are:

$(\frac{1}{2},\frac{1}{3},...\frac{1}{10}),(\frac{2}{1},\frac{2}{3},...,\frac{2}{10}),...,(\frac{10}{1},\frac{10}{2},\frac{10}{3},...,\frac{10}{9})$

So, total number of cases$=9\times 10=90$

$\therefore$ required probability$=\frac{17}{90}$