What is the Probability Distribution Function for Total Rotten Fruit?

[ouiouiwewe]

Hi guys,

Here's a description of the problem:

Suppose we have x apples and y oranges where each apple has a probability pa to rot and each orange has a probability po to rot, then what's the pdf of the total number of fruit z being rotten.

This pdf looks like a product of two binomial distributions on the surface. However, since z = x+y, x >= 0, y >= 0, then it is actually necessary to sum over all combinations of scenarios (i.e. 6 fruit rotten = 3 apples rotten + 3 oranges rotten or 6 fruits rotten = 1 apple rotten + 5 oranges rotten).

By intuition, I worked out the expectation to be x*pa + y *po and it appears to be correct when I manually tested my problem on a spreadsheet. However, I am not quite sure how that expectation can be derived from this messy pdf.

Any suggestions?

Thanks.

 

[Yayness]

The probability that z fruits will rot, is P(A)^0·P(O)^z+P(A)^1·P(O)^(z-1)+P(A)^2·P(O)^(z-2)+...+P(A)^z·P(O)^0

You could also write that as:
\sum^{z}_{k=0}P(A)^kP(O)^{z-k}

 

[ouiouiwewe]

This is close to the derivation I had, but instead of P(A)^[something] and P(O)^[something], I had two different binomial distributions. I have no problem getting the pdf, but I have trouble simplifying the expectation.

 

[Yayness]

Are you familiar with this division?
(a^n-b^n):(a-b)=a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+b^{n-1}

 

[ouiouiwewe]

I saw this sum in the derivation of the binomial expectation. Anyhow, this is what I have for the probability of having 2 rotten fruits:

prob of having 2 rotten apples * prob of having 0 rotten oranges
+
prob of having 1 rotten apple * prob of having 1 rotten orange
+
prob of having 0 rotten apples * prob of having 2 rotten oranges

This seems to be the product of two binomial distributions and then summed over all possible combinations.

 

[Yayness]

(Ignore my first post, there are errors in it.)

Let's say you have 2 fruits. Then:
-Both fruits can be apples. The probability for those being rotten is P(A)²
-Fruit 1 can be apple, fruit 2 can be orange. The probability for those being rotten is P(A)·P(O)
-Fruit 1 can be orange, fruit 2 can be apple. The probability for those being rotten is P(A)·P(O)
-Both fruits can be oranges. The probability for those being rotten is P(O)²

Since we don't know how many fruits we have, the probability that both the fruits will rot, is the average of the probabilities above, which is:
(P(A)²+2 P(A)·P(O)+P(O)²)/4

The probability that z fruits will be rotten is:
\frac{\binom{z}{0}P(A)^zP(O)^0+\binom{z}{1}P(A)^{z-1}P(O)^1+\binom{z}{2}P(A)^{z-2}P(O)^2+\ldots+\binom{z}{z}P(A)^0P(O)^z}{2^z}

This is exactly the same, but said in a shorter way:
\frac{\sum^{z}_{k=0}\binom{z}{k}P(A)^{z-k}P(O)^k}{2^z}

By using this equation: (a+b)^n=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\binom{n}{2}a^{n-2}b^2+\ldots+\binom{n}{n}a^0b^n

We get that:
\frac{\sum^{z}_{k=0}\binom{z}{k}P(A)^{z-k}P(O)^k}{2^z}=\frac{(P(A)+P(O))^z}{2^z}=\left(\frac{P(A)+P(O)}{2}\right)^z