Probability of Picking a Rotten Egg: Combinatorics Solution

[davon806]

Homework Statement

There are 20 eggs in the box and three of them are rotten,

If I pick eggs without replacement calculate P(my first rotten egg comes at my k-th pick)

Homework Equations

The Attempt at a Solution

I come up with two solutions:
(1)As there is no replacement,I can use the hypergeometric distribution
So,P(my first rotten egg comes at my k-th pick) = C(17,k-1)C(3,1) / C(20,k)

(2)I can use the multiplication theorem for conditional probability,i.e.
P(my first rotten egg comes at my k-th pick) = (17/20)(16/19)...(3/21-k)

Let assume I got a rotten egg in the 4-th trial,
Using method (1),P = 8/19
Using method (2),P = 2/19

Similarly,for the 5-th trial
(1) yields P = 35/76
(2) yields P = 7/76

In each case,(1)/(2) = k,where does this factor k come from?Both of them seems make sense so which one is wrong?

Thanks :)

 

[Samy_A]

Sorry, I was mistaken.

 

[davon806]

_______________________

 

[davon806]

I don't understand your second solution.
I do get the same result as your first solution.

Sorry for keep editing my post,my thought isn't clear.
Do you mean method(2)?
The probability of choosing good eggs out of 20 is 17/20
out of 19 is 16/19
.
.
.
Then suppose you got a rotten at the 4-th trial,
the probability would be (17/20)(16/19)(15/18)(3/17),because there are 3 rotten eggs out of 17 eggs
If we generalize the above ,we have
1 rotten egg = 3/20
2 rotten eggs = (17/20)(3/19)
3 rotten eggs = (17/20)(16/19)(3/18)
k rotten eggs = (17/20)...(3/21-k)

 

[Samy_A]

Yes, I was mistaken, sorry for that.

Your first method is wrong, because in the nominator you multiply the number of ways to select (k-1) non rotten eggs with 3 to get the number of positive cases. But that is clearly wrong, as the number of ways to get the first rotten egg at the k-th pick must be less that the number of ways to select (k-1) non rotten eggs.

Your second method seems right, although there is something wrong with the indices in the product (edit: you removed the product, so forget this remark ).

 

[davon806]

Yes, I was mistaken, sorry for that.

Your first method is wrong, because in the nominator you multiply the number of ways to select (k-1) non rotten eggs with 3 to get the number of positive cases. But that is clearly wrong, as the number of ways to get the first rotten egg at the k-th pick must be less that the number of ways to select (k-1) non rotten eggs.

Your second method seems right, although there is something wrong with the indices in the product (edit: you removed the product, so forget this remark ).

Do you mean the hypergeometric distribution would only make sense if we are choosing several objects in 1 trial?So in this case,in which the eggs are picked one by one,the hypergeometric is no longer applicable?

 

[Samy_A]

Do you mean the hypergeometric distribution would only make sense if we are choosing several objects in 1 trial?So in this case,in which the eggs are picked one by one,the hypergeometric is no longer applicable?

The hypergeometric distribution gives the probability of k successes in n draws (at what place in the picking order the successes occur is irrelevant).
That is not what is asked here: you are asked what the probability is that the k-th draw is the first success.

 

[Ray Vickson]

Sorry for keep editing my post,my thought isn't clear.
Do you mean method(2)?
The probability of choosing good eggs out of 20 is 17/20
out of 19 is 16/19
.
.
.
Then suppose you got a rotten at the 4-th trial,
the probability would be (17/20)(16/19)(15/18)(3/17),because there are 3 rotten eggs out of 17 eggs
If we generalize the above ,we have
1 rotten egg = 3/20
2 rotten eggs = (17/20)(3/19)
3 rotten eggs = (17/20)(16/19)(3/18)
k rotten eggs = (17/20)...(3/21-k)

Your calculations are correct, but are severely mis-labelled. You have three rotten eggs---period. You are asked to find the probability of encountering your first rotten egg on draw k, so if $X =$ draw number of first rotten egg, then you should, indeed, have
$$P(X = 1) = \frac{3}{20}, \: P(X = 2) = \frac{17}{20} \frac{3}{19}, \; \text{etc.}$$

You could have labelled these instead as P(0 good eggs) = 3/20, P(1 good egg) = (17/20)(3/19), etc., because you would be counting the "good" eggs before your first rotten one.

BTW: this is not a "combinatorics" problem; it is a "probability" problem.

 

[davon806]

Thx for reminding, I always have a problem on defining the RV(i.e.X=1,X=2) so I usually just do it in that way, though it is one of the culprits lead to confusions. I will write them more precisely.