# Probability of trees growing

#### cpaiz

##### New member
Hello,
I'm new to this site and i am having trouble with probability questions.
A man plants ten trees and the probability that each tree will grow is .45. What is the probability that all ten trees grow?
-is the answer just .45^10?
Also, what is the probability that at least two trees grow?
-.45^2?
If X is the random variable where the value of X is the number of trees that grow, find the mean and variance of X?
-i have no idea how to figure this out. help please. =)

#### tkhunny

##### Moderator
Staff member
First, we need an assumption of Independence. Without this, we have troubles. For example, if one tree grows REALYL QUICKLY and blocks the sun so no others can grow, clearly they are NOT independent.

Assuming independence.

Pr(all 10) = 0.45^10 -- Very good.

Pr(at least two) You have Pr(two out of two will grow). this si no good.
It is Binomial. Expand (0.45 + 0.55)^{10} and pick the terms with 0.45^10, 0.45^9, 0.45 ^8, ...0.45^2
Actyually, I'm kidding a little. Subtract from unity (1) the sum of the terms with 0.55^0 and 0.55^1.

Variance. Again, it;s binomial. There are formulas for this. do you have them?

#### cpaiz

##### New member
Pr(two trees out of ten will grow): i am still confused how you came up with that answer.
Also, i do have the formulas for mean and variance, but what numbers am i supposed to plug in from the problem?
Thank you. This is an online course i am taking so its hard to understand the material on my own.

#### tkhunny

##### Moderator
Staff member
Your outline had better tell you how to calculate a binomial coefficient, or really, you cannot be expected to complete problems of this type.

If p is the probability of success,
The q = 1-p is the probability of failure.

You have p = 0.45 and q = 1-p = 0.55

You also have n = 10, ten trees.

$$\displaystyle Pr(exactly\; ten\; succeeding) = \frac{10!}{10!0!}p^{10}q^{0}$$

$$\displaystyle Pr(exactly\; nine\; succeeding) = \frac{10!}{9!1!}p^{9}q^{1}$$

$$\displaystyle Pr(exactly\; eight\; succeeding) = \frac{10!}{8!2!}p^{8}q^{2}$$

etc...

These are the same things you would produce if you went to all the algebra trouble to expand $$\displaystyle (p+q)^{10}$$.

You must find this section of your materials and get familiar with it.

Again, the easiest way to solve this "at least two" question is to calculate Pr(0) + Pr(1) and subtract the result from unity (1).

You should have Mean = np and Variance = npq.