Calculating Binomial Probabilities
Add yours
Sorry, we don't have any videos for this topic yet.
Let teachers know you need one by requesting it
Key Questions

Answer:
See below on my thoughts:
Explanation:
The general form for a binomial probability is:
#sum_(k=0)^(n)C_(n,k)(p)^k((~p)^(nk))# The question is Why do we need that first term, the combination term?
Let's work an example and then it'll come clear.
Let's look at the binomial probability of flipping a coin 3 times. Let's set getting heads to be
#p# and of not getting heads#~p# (both#=1/2)# .When we go through the summation process, the 4 terms of the summation will equal 1 (in essence, we are finding all the possible outcomes and so the probability of all the outcomes summed up is 1):
#sum_(k=0)^(3)=color(red)(C_(3,0)(1/2)^0((1/2)^(3)))+color(blue)(C_(3,1)(1/2)^1((1/2)^(2)))+C_(3,2)(1/2)^2((1/2)^(1))+C_(3,3)(1/2)^3((1/2)^(0))# So let's talk about the red term and the blue term.
The red term describes the results of getting 3 tails. There is only 1 way for that to be achieved, and so we have a combination that equals 1.
Note that the last term, the one describing getting all heads, also has a combination that equals 1 because again there is only one way to achieve it.
The blue term describes the results of getting 2 tails and 1 head. There are 3 ways that can happen: TTH, THT, HTT. And so we have a combination that equals 3.
Note that the third term describes getting 1 tails and 2 heads and again there are 3 ways to achieve that and so the combination equals 3.
In fact, in any binomial distribution, we have to find the probability of a single kind of event, such as the probability of achieving 2 heads and 1 tails, and then multiplying it by the number of ways it can be achieved. Since we don't care about the order in which the results are achieved, we use a combination formula (and not, say, a permutation formula).

This key question hasn't been answered yet. Answer question

This key question hasn't been answered yet. Answer question