From a heap containing 15 pairs of shoes,10 shoes are selected at random. The probability that there is no complete pair in the selected shoes is
$
{\text{A}}{\text{. }}\dfrac{{^{30}{{\text{C}}_{10}} - {(^{15}}{{\text{C}}_{10}}){2^{10}}}}{{^{30}{{\text{C}}_{10}}}} \\
{\text{B}}{\text{. }}\dfrac{{^{15}{{\text{C}}_{10}}({2^{10}})}}{{^{30}{{\text{C}}_{10}}}} \\
{\text{C}}{\text{. }}\dfrac{{^{30}{{\text{C}}_{10}} - {2^{15}}}}{{^{30}{{\text{C}}_{10}}}} \\
{\text{D}}{\text{. }}\dfrac{{^{15}{{\text{C}}_{10}}}}{{^{30}{{\text{C}}_{10}}}} \\
$
Answer
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Hint:-Use the concept of both probability and combinations. The probability is the ratio of favourable outcome to the total number of outcomes.
Given, a heap is containing 15 pairs of shoes. We need to find the probability of selecting 10 shoes at random with no complete pair.
So, total numbers of shoes are 15(2) i.e. 30 shoes.
Probability is the ratio of number of ways possible to the total number of outcomes.
i.e. Probability = $\dfrac{{{\text{no}}{\text{. of ways possible}}}}{{{\text{total no}}{\text{. of outcome}}}}$ --(1)
For selecting the shoes from a heap, we need to know the concept of combination.
A combination is selection of all or part of a set of objects, without regard to order of selection. It is represented as $^{\text{n}}{{\text{C}}_{\text{r}}}$ where n is the number of total objects and r is the number of objects to be selected.
And, $^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{{\text{r}}!\left( {{\text{n}} - {\text{r}}} \right)!}}$ --(2)
Where n! =${\text{n}} \times {\text{(n - 1)}} \times {\text{(n - 2)}}...{\text{2}} \times {\text{1}}$ and similarly r! = ${\text{r}} \times {\text{(r - 1)}} \times {\text{(r - 2)}}...{\text{2}} \times {\text{1}}$
Now, for finding the probability we need to find the number of ways the selection of 10 shoes is possible.
No. of ways possible for selection of 10 pairs randomly from 15 pairs is $^{15}{{\text{C}}_{10}}$ and for selecting a shoe from each pair can be given by ${{(^2}{{\text{C}}_1})^{}}$. Since, we have sorted 10 pairs. So, we need to select shoes from each pair . So, total possibility of selection of 10 shoe from 10 pairs will be ${{(^2}{{\text{C}}_1})^{10}}$
So, number of ways possible = $^{15}{{\text{C}}_{10}} \times $${{(^2}{{\text{C}}_1})^{10}}$
Total number of possible outcomes = selecting 10 shoes randomly from 30 shoes
=$^{30}{{\text{C}}_{10}}$
Putting both the value in equation (1) we get
Probability = $\dfrac{{^{15}{{\text{C}}_{10}} \times {{{(^2}{{\text{C}}_1})}^{10}}}}{{^{30}{{\text{C}}_{10}}}}$ --(3)
Now, ${{(^2}{{\text{C}}_1})^{}}$ = $\dfrac{{2!}}{{1!\left( {2 - 1} \right)!}}$=$\dfrac{{2 \times 1}}{{1 \times 1}}$= 2
Putting the value of ${{(^2}{{\text{C}}_1})^{}}$in the equation (3) , we get
Probability = $\dfrac{{^{15}{{\text{C}}_{10}} \times {{(2)}^{10}}}}{{^{30}{{\text{C}}_{10}}}}$
Hence, option (b) is the correct answer.
Note:- In these types of questions , we need to remember the concept of both probability and combinations. While applying a combination formula, we need to remember that a similar entity’s count can be placed in the formula i.e. if n = total no. of shoe’s pair then r = no. of selected pair of shoes or if n = total no. of shoes then r = no. of selected shoes.
Given, a heap is containing 15 pairs of shoes. We need to find the probability of selecting 10 shoes at random with no complete pair.
So, total numbers of shoes are 15(2) i.e. 30 shoes.
Probability is the ratio of number of ways possible to the total number of outcomes.
i.e. Probability = $\dfrac{{{\text{no}}{\text{. of ways possible}}}}{{{\text{total no}}{\text{. of outcome}}}}$ --(1)
For selecting the shoes from a heap, we need to know the concept of combination.
A combination is selection of all or part of a set of objects, without regard to order of selection. It is represented as $^{\text{n}}{{\text{C}}_{\text{r}}}$ where n is the number of total objects and r is the number of objects to be selected.
And, $^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{{\text{r}}!\left( {{\text{n}} - {\text{r}}} \right)!}}$ --(2)
Where n! =${\text{n}} \times {\text{(n - 1)}} \times {\text{(n - 2)}}...{\text{2}} \times {\text{1}}$ and similarly r! = ${\text{r}} \times {\text{(r - 1)}} \times {\text{(r - 2)}}...{\text{2}} \times {\text{1}}$
Now, for finding the probability we need to find the number of ways the selection of 10 shoes is possible.
No. of ways possible for selection of 10 pairs randomly from 15 pairs is $^{15}{{\text{C}}_{10}}$ and for selecting a shoe from each pair can be given by ${{(^2}{{\text{C}}_1})^{}}$. Since, we have sorted 10 pairs. So, we need to select shoes from each pair . So, total possibility of selection of 10 shoe from 10 pairs will be ${{(^2}{{\text{C}}_1})^{10}}$
So, number of ways possible = $^{15}{{\text{C}}_{10}} \times $${{(^2}{{\text{C}}_1})^{10}}$
Total number of possible outcomes = selecting 10 shoes randomly from 30 shoes
=$^{30}{{\text{C}}_{10}}$
Putting both the value in equation (1) we get
Probability = $\dfrac{{^{15}{{\text{C}}_{10}} \times {{{(^2}{{\text{C}}_1})}^{10}}}}{{^{30}{{\text{C}}_{10}}}}$ --(3)
Now, ${{(^2}{{\text{C}}_1})^{}}$ = $\dfrac{{2!}}{{1!\left( {2 - 1} \right)!}}$=$\dfrac{{2 \times 1}}{{1 \times 1}}$= 2
Putting the value of ${{(^2}{{\text{C}}_1})^{}}$in the equation (3) , we get
Probability = $\dfrac{{^{15}{{\text{C}}_{10}} \times {{(2)}^{10}}}}{{^{30}{{\text{C}}_{10}}}}$
Hence, option (b) is the correct answer.
Note:- In these types of questions , we need to remember the concept of both probability and combinations. While applying a combination formula, we need to remember that a similar entity’s count can be placed in the formula i.e. if n = total no. of shoe’s pair then r = no. of selected pair of shoes or if n = total no. of shoes then r = no. of selected shoes.
Last updated date: 19th Sep 2023
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