Question

Two numbers ‘a’ and ‘b’ are selected successively without replacement in that order from the integers 1 to 10. The probability that $\dfrac{a}{b}$ is an integer is:
$\left( a \right)\dfrac{17}{45}$
$\left( b \right)\dfrac{1}{5}$
$\left( c \right)\dfrac{17}{90}$
$\left( d \right)\dfrac{8}{45}$

Answer 1

To solve this question, we will, first of all, compute all the cases such that $\dfrac{a}{b}$ is an integer by calculating so. Now, we will compute the total cases possible. After that, we will determine the probability by using the formula, $\text{Probability}=\dfrac{\text{Favorable Outcomes}}{\text{Total Outcomes}}.$

Complete step-by-step answer:
Given that a and b are selected without replacement from 1 to 10. Now, we will consider all favorable cases for which $\dfrac{a}{b}$ will be an integer. The following are consecutive $\dfrac{a}{b}$ from an integer.
$\dfrac{2}{1}=2;\dfrac{3}{1}=3;\dfrac{4}{1}=4;\dfrac{4}{2}=2;\dfrac{5}{1}=5;\dfrac{6}{1}=6;\dfrac{6}{2}=3;\dfrac{6}{3}=2;\dfrac{7}{1}=7;\dfrac{8}{1}=8;\dfrac{8}{2}=4;\dfrac{8}{4}=2;\dfrac{9}{1}=9;\dfrac{10}{1}=10;\dfrac{9}{3}=3;\dfrac{10}{2}=5;\dfrac{10}{5}=2$
So, counting all the given above numbers, we have the total numbers of cases as 17. The total cases are to be determined now. The total cases can be determined by writing all the possibilities of the number from 1 to 10. The total cases are
$\left( \dfrac{1}{2},\dfrac{1}{3},...\dfrac{1}{10} \right),\left( \dfrac{2}{1},\dfrac{2}{3},\dfrac{2}{4}.....\dfrac{2}{10} \right),\left( \dfrac{10}{1},.....\dfrac{10}{9} \right)$
We have 10 brackets and every bracket has 9 elements in it.
Therefore, the total number of cases are $9\times 10=90.$
The probability can be determined by using the formula $\text{Probability}=\dfrac{\text{Favorable Outcomes}}{\text{Total Outcomes}}.$
Here, the favorable outcomes are 17 and the total number of outcomes are 90. Therefore we get,
$\Rightarrow \text{Required Probability}=\dfrac{17}{90}$

So, the correct answer is “Option (c)”.

Note: The key point to note here is that we will consider the total cases as
$\left( \dfrac{1}{2},\dfrac{1}{3},...\dfrac{1}{10} \right),\left( \dfrac{2}{1},\dfrac{2}{3},\dfrac{2}{4}.....\dfrac{2}{10} \right),\left( \dfrac{10}{1},.....\dfrac{10}{9} \right)$
We have not considered the parts where $\dfrac{1}{1},\dfrac{2}{2},\dfrac{3}{3}....\dfrac{10}{10}$ is there. This is because $\dfrac{a}{b}$ is used and a and b are different numbers, also they are taken without replacement.