1. ## Combination of events

Hello forumites
Ten pairs of shoes are in a closet. Four shoes are selected at random. Author wants me to find the probability that there will be at least one pairs of shoes among the four shoes selected.
Solution:-

Four shoes can be selected out of 10 pairs (20 number) in $\binom{20}{4}$ ways. Now we want to find the probability that there will be at least one pair of shoes among the four shoes selected which is equal to the probability that remains after deducting the probability of no pairs of shoes among the four shoes selected from the total probability.. So it is $1-\frac{\binom{10}{4}}{\binom{20}{4}}=0.956656$
But answer provided is $\frac{99}{323}=\frac{\binom{55}{2}}{\binom{20}{4} }$ Now which is wrong?

2. ## Re: Combination of events

If shoes within a pair are indistinguishable I agree with your answer.

I can't imagine where they came up with $\dbinom{55}{2}$

3. ## Re: Combination of events

Originally Posted by romsek
If shoes within a pair are indistinguishable I agree with your answer.

I can't imagine where they came up with $\dbinom{55}{2}$
Ok I was wrong. They are correct. Though $\dbinom{55}{2}$ is just coincidence I believe

Let's pick shoes so that there are no pairs

first pick is 20/20 shoes.
second pick is 18/19
third is 16/18
fourth is 14/17

We want 1 minus the product of these

$p = 1 - \dfrac{20 \cdot 18 \cdot 16 \cdot 14}{20 \cdot 19 \cdot 18 \cdot 17} = 1 - \dfrac{224}{323} = \dfrac{99}{323}$

4. ## Re: Combination of events

Originally Posted by romsek
Ok I was wrong. They are correct. Though $\dbinom{55}{2}$ is just coincidence I believe

Let's pick shoes so that there are no pairs

first pick is 20/20 shoes.
second pick is 18/19
third is 16/18
fourth is 14/17

We want 1 minus the product of these

$p = 1 - \dfrac{20 \cdot 18 \cdot 16 \cdot 14}{20 \cdot 19 \cdot 18 \cdot 17} = 1 - \dfrac{224}{323} = \dfrac{99}{323}$
Hello,

5. ## Re: Combination of events

Originally Posted by Vinod
Hello,
The problem with your answer is that by picking pairs, you remove the shoe in each pair that you don't pick from the possible shoes to pick.

6. ## Re: Combination of events

Originally Posted by romsek
The problem with your answer is that by picking pairs, you remove the shoe in each pair that you don't pick from the possible shoes to pick.
Hello,
If we want to compute the probability of exactly one pair among the four selected from the total ten pairs of shoes, how it should be computed? Secondly what is the probability of two pairs among the four selected out of ten pairs of shoes?

7. ## Re: Combination of events

Originally Posted by Vinod
Hello,
If we want to compute the probability of exactly one pair among the four selected from the total ten pairs of shoes, how it should be computed?
do you mean by randomly selecting pairs of shoes or by randomly selecting single shoes?

8. ## Re: Combination of events

Originally Posted by romsek
do you mean by randomly selecting pairs of shoes or by randomly selecting single shoes?
Hello,
It means random selection of four single shoes out of twenty shoes which are equal to ten pairs of shoes.

9. ## Re: Combination of events

Originally Posted by romsek
do you mean by randomly selecting pairs of shoes or by randomly selecting single shoes?
$p = \dfrac{20 \cdot 1 \cdot 18 \cdot 16}{20\cdot 19 \cdot 18 \cdot 17} = \dfrac{16}{323}$

10. ## Re: Combination of events

Originally Posted by romsek
$p = \dfrac{20 \cdot 1 \cdot 18 \cdot 16}{20\cdot 19 \cdot 18 \cdot 17} = \dfrac{16}{323}$
Hello,

11. ## Re: Combination of events

From the ten pairs of shoes, choose 3. Among the 3, choose one pair to take both shoes. From the other two pairs, choose one of the two for each. The total number of ways to select the shoes:

$$\dbinom{10}{3}\dbinom{3}{1}\dbinom{2}{1}\dbinom{ 2}{1}$$

Out of the total of 20 shoes where you choose 4.

$$\dbinom{20}{4}$$

So, the probability is:

$$\dfrac{\dbinom{10}{3}\dbinom{3}{1}\dbinom{2}{1} \dbinom{2}{1}}{\dbinom{20}{4}} = \dfrac{96}{323}$$

12. ## Re: Combination of events

Originally Posted by Vinod
Hello forumites
If an insect forum exists, I will greet them as forum mites.