1. ## Statisics

If a seed is planted, it has a 70% chance of growing into a healthy plant.
If 8 seeds are planted, what is the probability that exactly 1 doesn't grow?

2. ## Re: Statisics

The number of healthy plants has a binomial distribution with $n=8,~p=0.7$

$P[\text{1 plant doesn't grow}] = P[\text{7 plants grow}] = \dbinom{8}{7}(0.7)^7(1-0.7)^1 = \dfrac{2470629}{12500000} = 0.19765$

3. ## Re: Statisics

Writing "G" for "grows" and "N" for "does not grow", this can happen as
NGGGGGGG
GNGGGGGG
GGNGGGGG
GGGNGGGG
GGGGNGGG
GGGGGNGG
GGGGGGNG
GGGGGGGN

In other words, there are 8 possible orders. That is reflected in Romsek's $\displaystyle \begin{pmatrix} 8 \\ 7\end{pmatrix}= \frac{8!}{7!1!}= 8$. The probability of "NGGGGGGG" is, of course, $\displaystyle (.3)(.7)^7= 0.02471$, approximately, and it is easy to see that the probability of "GNGGGGGG", etc is exactly the same- you have the same numbers, just in a different order. So the probability of all 8 orders is 8(0.02471)= 0.19768.