generals336

2021-08-21

A closet contains 10 pairs of shoes.
If 8 shoes are randomly selected, what is the probability that there will be
(a) no complete pair?
(b) exactly 1 complete pair?

Velsenw

(a)
Since there shouldn’t be a complete pair, we are allowed to choose at most 1 shoe from each pair.
Therefore, our choice can be made in the following way: first choose 8 pairs from the 10 pairs in the closet. this can be done in $\left(\begin{array}{c}10\\ 8\end{array}\right)$ ways.
Then from chosen 8 pairs, choose one shoe from each.
This can be done in ${2}^{8}$ ways as there are 2 choices for each pair.
Therefore, total number of choices is $\left(\begin{array}{c}10\\ 8\end{array}\right){2}^{8}.$.
The number of ways to randomly choose 8 shoes out of 20 is $\left(\begin{array}{c}20\\ 8\end{array}\right)$. Therefore,
$P\left(\text{no complete pair}\right)=\frac{\left(\left(10\right),\left(8\right)\right){2}^{8}}{\left(\left(20\right),\left(8\right)\right)}$
(b)
For getting exactly one pair, we can first choose the pair which will appear completely. there are 10 ways of doin it.
Then we need to choose 6 shoes from 9 remaining pairs.
Proceeding as part(a) we see that there are ((9),(6))2^6 ways for this. Therefore, the total number is $\left(\begin{array}{c}9\\ 6\end{array}\right){2}^{6}\cdot 10$. Thus,
$P\left(\text{exactly one pair}\right)=\frac{\left(\left(9\right),\left(6\right)\right){2}^{6}\ast 10}{\left(\left(20\right),\left(8\right)\right)}$
Result: $P\left(\text{no complete pair}\right)=\frac{\left(\left(10\right),\left(8\right)\right){2}^{8}}{\left(\left(20\right),\left(8\right)\right)}$
$P\left(\text{exactly one pair}\right)=\frac{\left(\left(9\right),\left(6\right)\right){2}^{6}\ast 10}{\left(\left(20\right),\left(8\right)\right)}$

star233

Step 1:
(a) To find the probability of selecting no complete pair when 8 shoes are randomly selected, we need to calculate the number of favorable outcomes (no complete pair) and divide it by the total number of possible outcomes.
The total number of ways to select 8 shoes out of 20 (10 pairs) is given by $\left(\genfrac{}{}{0}{}{20}{8}\right)$.
The number of ways to select 8 shoes with no complete pair can be calculated as follows:
- Choose 8 shoes from 10 pairs: $\left(\genfrac{}{}{0}{}{10}{4}\right)$
- Multiply by 2 for each pair (since we can choose either the left or right shoe from each pair): ${2}^{4}$
Therefore, the number of favorable outcomes is $\left(\genfrac{}{}{0}{}{10}{4}\right)·{2}^{4}$.
The probability of no complete pair is given by:

Step 2:
(b) To find the probability of exactly 1 complete pair when 8 shoes are randomly selected, we again need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
The number of ways to select 1 complete pair from the 10 available pairs is $\left(\genfrac{}{}{0}{}{10}{1}\right)$.
Once we have selected a complete pair, we need to choose 6 more shoes from the remaining 18 shoes (9 pairs minus the selected pair). This can be done in $\left(\genfrac{}{}{0}{}{18}{6}\right)$ ways.
For each selected pair, we can choose the left or right shoe, so we multiply by 2.
Therefore, the number of favorable outcomes is $\left(\genfrac{}{}{0}{}{10}{1}\right)·\left(\genfrac{}{}{0}{}{18}{6}\right)·2$.
The probability of exactly 1 complete pair is given by:

alenahelenash

(a) To find the probability that there will be no complete pair when 8 shoes are randomly selected from 10 pairs, we need to calculate the number of ways this can happen and divide it by the total number of possible outcomes.
The total number of ways to select 8 shoes from 20 (10 pairs) is given by the binomial coefficient:
$\left(\genfrac{}{}{0}{}{20}{8}\right)$
The number of ways to select 8 shoes without any complete pair is the same as selecting 8 shoes from 18 individual shoes (9 pairs minus 1 complete pair), which can be calculated as:
$\left(\genfrac{}{}{0}{}{18}{8}\right)$
Therefore, the probability of no complete pair is:

(b) To find the probability of exactly 1 complete pair, we need to consider two cases:
Case 1: Selecting 1 complete pair and 6 individual shoes.
The number of ways to select 1 complete pair is:
$\left(\genfrac{}{}{0}{}{10}{1}\right)$
And the number of ways to select 6 individual shoes from the remaining 18 is:
$\left(\genfrac{}{}{0}{}{18}{6}\right)$
Case 2: Selecting 2 complete pairs and 4 individual shoes.
The number of ways to select 2 complete pairs is:
$\left(\genfrac{}{}{0}{}{10}{2}\right)$
And the number of ways to select 4 individual shoes from the remaining 16 is:
$\left(\genfrac{}{}{0}{}{16}{4}\right)$
The total number of favorable outcomes for exactly 1 complete pair is the sum of the two cases:

Therefore, the probability of exactly 1 complete pair is:

Do you have a similar question?