Solving a Probability Problem with 20 Chickens

[Ajoo]

Hi, i some time ago my physics teacher asked me this problem: There are 20 chickens coming out of an hen-coop. 15 are white, 5 are black. What's the probability of at least 2 black chickens come out consecutively.

so basically i tried to calc P = 1 - (Probability of no black chicken come out consecutively )

So i thought P = 1 - D(20,5)/C(20,5)

being D = Ways to arrange the 5 chickens in 20 "boxes" so that the 5 black never come out consecutivelly

after some thinking i figured out that
D(n,p) = D(n-2,p-1) + D(n-1,p)

which is somewhat similar to C(n,p) = C(n-1,p-1) + C(n-1,p)

So i started making some sort of pascal triangle for the D function which lead me to this complicated sum:

$$D(n,p) = \sum_{i=1}^{n-2p+2} C(p-3+i,p-2) \times (k-i+1)$$

hope i got this right (first time using latex ^^)

well, i came to the same result as my teacher but he had a much more simple formula, to which he arrived empirically:

P = 1 - C(20-5+1,5)*15!*5!/20!

So, D(n,p)=C(n-p+1,p) i tried to deduce this formula from what i previously had but with no success. Can some1 please tell me how i could have arrived to it.

Ty

PS: By the way, I'm only in high school so try to keep it simple.

 

[NateTG]

Here's one way:

Saying that there are no two black chickens in a row is the same as saying that every black chicken is directly in front of a white chicken, or directly in front of the end of the row.

So we can think of this as 16 spots (1 in front of each white chicken and 1 at the end) and choosing 5.

You should be able to generalize from there.

 

[Gokul43201]

How many ways are there of putting k black objects in the n+1 spaces formed by a string of n white objects, with no more than 1 black object per space ?

Edit : This seems to be exactly what Nate was saying. The spots threw me! I guess I'm a space person.