Calculate the probability of k successes for a binomial experiment consisting of n trials with the probability p of success on each trial: k ≥, n=5, p=0.2

Answer 1

Probability of k success in a binomial distribution is calculated taking two things into account:

1) The amount of ways you can k success can occur in n attempts/trials.

Lets say our k=1, meaning we are only success 1 time, and our n=3, meaning we have attempted our trail 3 times. That means there are 3 ways this could have happened: a) S,F,F; b) F,S,F; c) F,F,S. Each of these order of events are different and have there own chance of happening. However, counting these combination can become difficult when experiments become larger.

Formula developed for counting the number of ways you can have k success in n trails:

n!/(k!*(n-k)!

For my example above: 3!/(1!*(3-1)!) = 3

2) The probability of a certain event occurring

This is where we look at p, the frequency which we except to see a success full attempt in each individual n trails. In a binomial experiment, each one of these n trials are independent. This means we can multiply the the probability of each success, p, and the probability of each failure, 1-p, to find the probability of each individual order of success and failure.

Using my previous example, the probability of:

a) = p*(1-p)*(1-p) -> p*(1p)^(2)

b) = (1-p)*p*(1-p) -> p*(1p)^(2)

c) = (1-p)*(1-p)*p -> p*(1p)^(2)

Notice how each event with the same number of success in a binomial experiment has the same result after multiplying the probability of each trial!

With these two pieces of information, you can then multiply the probability of having k success by the amount of ways you can see k success happen in your binomial experiment because every way k success can happen in you binomial experiment have the same probability of occurring. The result of this multiplication is your probability of k success happening in your binomial experiment.

Looking at the example provided n=5, p=0.2, and k is not given. Let's assume k is 2.

1) How many different ways can 2 success occur in 5 trials?

5!/(2!*(5-2)!) = 10

2) What is the probability for any one of the different ways I can 2 success occur in 5 trails?

p*p*(1-p)*(1-p)*(1-p) = 0.2*0.2*0.8*0.8*0.8 = (0.2)^(2)*(0.8)^(3) = 0.02048

3) What probability of 2 successes happening during 5 trials when the probability of a single success is 0.2?

10*0.02048 = 0.2048