Probability to pick a specific element

[aarocorp]

Morning,
With my colleagues we cannot agree on who is right, so we are asking for the community help to us to get some light.

So there is the problem:

We have a basket of 26 unique elements (lets say from a to z).
What is the probability to get the "e" element at the 5th draw.

Note : Once picked the element is not put back into the basket and we don't know what are the previous elements drawn

thanks in advance for your help

 

[Romsek]

On the first draw we can select 25 of the 26 elements (i.e. not "e")
Similar for draws 2-4, 24 of 25, 23 of 24, etc.

On the 5th draw we must draw the "e", i.e. 1 out of the remaining 22 elements
Having drawn the "e" on the 5th draw we can subsequently choose anything.

Given all this can you now assemble the factors making up the probability?
It will be the product of the probability of acceptable picks, for draws 1-5.

 

[Deleted member 4993]

Morning,
With my colleagues we cannot agree on who is right, so we are asking for the community help to us to get some light.

So there is the problem:

We have a basket of 26 unique elements (lets say from a to z).
What is the probability to get the "e" element at the 5th draw.

Note : Once picked the element is not put back into the basket and we don't know what are the previous elements drawn

thanks in advance for your help

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment
Hint:

Probability of NOT picking E on draw #1 = 25/26

Probability of NOT picking E on draw #2 = 24/25

Probability of NOT picking E on draw #3 = 23/24

Probability of NOT picking E on draw #4 = 22/23

Probability of picking E on draw #5 = 1/22

Now put ALL these together.....

 

[aarocorp]

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this assignment
Hint:

Probability of NOT picking E on draw #1 = 25/26

Probability of NOT picking E on draw #2 = 24/25

Probability of NOT picking E on draw #3 = 23/24

Probability of NOT picking E on draw #4 = 22/23

Probability of picking E on draw #5 = 1/22

Now put ALL these together.....

We haven't shared my POV because we didn't want to influence the answers we might get (to keep them neutral).
Sorry about that ;-)

 

[aarocorp]

On the first draw we can select 25 of the 26 elements (i.e. not "e")
Similar for draws 2-4, 24 of 25, 23 of 24, etc.

On the 5th draw we must draw the "e", i.e. 1 out of the remaining 22 elements
Having drawn the "e" on the 5th draw we can subsequently choose anything.

Given all this can you now assemble the factors making up the probability?
It will be the product of the probability of acceptable picks, for draws 1-5.

Those were the answers we had :
1st : one colleague said the probability stay 1/26 no matter how many draws are made
2nd : it is a conditional probability of picking "e" knowing "4 draws" have been made (but he can't really make the calculations either)

So we actually don't know how to mathematically compute this probability

 

[pka]

So there is the problem:
We have a basket of 26 unique elements (lets say from a to z).
What is the probability to get the "e" element at the 5th draw.

There are \(\displaystyle \mathscr{P}^{26}_5=\dfrac{26!}{(26-5)!} =7893600\) ordered ways to draw five from the basket.
Of those there are \(\displaystyle \mathscr{P}^{25}_4=\dfrac{25!}{(25-4)!} =303600\) ordered ways to draw five having \(\displaystyle 5\) in the fifth position.
That gives the probability of \(\displaystyle \frac{303600}{7893600}=\frac{1}{26}\).

 

[Romsek]

As you've now been told the answer let's confirm the method outlined in the first couple of posts agrees.

\(\displaystyle p = \dfrac{25\cdot 24\cdot 23\cdot 22 \cdot 1}{26\cdot 25\cdot 24\cdot 23 \cdot 22} = \dfrac{1}{26}\)

 

[aarocorp]

Thank you all for the answer. The problem is now solved