Question

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

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

Answer 1

The correct option is C

$\frac{17}{90}$

Integer:

A whole number that can be zero, positive, or negative is called an integer.

Probability:

Probability is defined as the possibility of a random event occurring.

Calculating the probability of getting an integer:

The variables a and b can take values from $1$ to $10$.

Consider S to be the sample space.

The possible outcomes in the form $\frac{\mathrm{a}}{\mathrm{b}}$ without replacement are $\left(\frac{1}{2},\frac{1}{3},\frac{1}{4},...,\frac{1}{10}\right),\left(\frac{2}{1},\frac{2}{3},\frac{2}{4},...,\frac{2}{10}\right),...,\left(\frac{10}{1},\frac{10}{2},\frac{10}{3},...,\frac{10}{9}\right)$.

Total number of possible outcomes, $\mathrm{n}\left(\mathrm{S}\right)=9\times 10=90$

Consider A to be an event that $\frac{\mathrm{a}}{\mathrm{b}}$ is an integer.

The favorable outcomes such that $\frac{\mathrm{a}}{\mathrm{b}}$ is an integer without replacement are $\frac{2}{1},\frac{3}{1},\frac{4}{1},\frac{4}{2},\frac{5}{1},\frac{6}{1},\frac{6}{2},\frac{6}{3},\frac{7}{1},\frac{8}{1},\frac{8}{2},\frac{8}{4},\frac{9}{1},\frac{9}{3},\frac{10}{1},\frac{10}{2},\frac{10}{5}$.

Number of favorable outcomes, $\mathrm{n}\left(\mathrm{A}\right)=17$

$\mathrm{P}\left(\mathrm{A}\right)$ is the probability that $\frac{\mathrm{a}}{\mathrm{b}}$is an integer.

$$\begin{gathered} \Rightarrow \mathrm{P}\left(\mathrm{A}\right)=\frac{\mathrm{n}\left(\mathrm{A}\right)}{\mathrm{n}\left(\mathrm{S}\right)} \\ \Rightarrow \mathrm{P}\left(\mathrm{A}\right)=\frac{17}{90} \end{gathered}$$

Therefore the probability of getting an integer in the form $\frac{\mathbf{a}}{\mathbf{b}}$ is $\frac{\mathbf{17}}{\mathbf{90}}$.