Combination Feature of Probability Question We are given the following scenario:
A nursery claims that 80% of its trees survive at least three years. As part of a service learning requirement, you are landscaping a neighborhood park and have been asked to plant four trees. You wonder what the probability is that all four will survive at least three years.
Q: Using your calculator combination feature and exponential capabilities to find the probability of all 4 trees survived.
I wasn't sure how to use my calculator for this question but wouldn't you just did .8*.8*.8*.8  since we have an 80% survival rate and these events are all independent of one another.  Even though I think my math logic is correct can somebody explain to me what the combination feature is and what this would look like plugging in this question into the calculator.   
 A: As mentioned, we can deduce that if $\mathbb{X}$ is a discrete random variable representing the number of trees, then $\mathbb{X}$ has a binomial distribution (specifically $\mathbb{X}\sim(n=4,p=0.8))$.
Recall the probability mass function for the binomial distribution $\mathbb{X}\sim(n=,p)$, where n is the sample size and p is the probability of success, is $\mathbb{P}(\mathbb{X}=x)={n\choose x}p^xq^{n-x}$, given that $q=1-p$.
As you surmised, this is indeed equal to $0.8^4=0.4096$.
Specifically, since we are asked to find the probability that exactly four trees survive, we calculate $\mathbb{P}(\mathbb{X}=4)=p(4)={4\choose 4}(0.8^4)(0.2^0)=(1)(0.8^4)(1)=0.8^4=0.4096$.
A: 
Even though I think my math logic is correct can somebody explain to me what the combination feature is and what this would look like plugging in this question into the calculator.

The combinatorial factor of the answer would be ${^4\mathsf C_4}$, which equals one, and so...
$${^4\mathsf C_4}\,(0.8)^4~~=~~(0.8)^4$$

This is not the best chosen question for demonstrating the combinatoric feature in calculations.
