# Product of two binomials

by ouiouiwewe
Tags: binomials, product
 P: 38 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: $$\sum^{z}_{k=0}P(A)^kP(O)^{z-k}$$
 P: 38 Product of two binomials Are you familiar with this division? $$(a^n-b^n):(a-b)=a^{n-1}+a^{n-2}b+a^{n-3}b^2+\ldots+b^{n-1}$$
 P: 38 (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: $$\frac{\sum^{z}_{k=0}\binom{z}{k}P(A)^{z-k}P(O)^k}{2^z}=\frac{(P(A)+P(O))^z}{2^z}=\left(\frac{P(A)+P(O)}{2}\righ t)^z$$