Bernoulli trial

Question

At least 2 failures in 9 trials with p = 0.8 

Answer 1

We assume this is a Bernoulli question - it is not.

This is a Binomial question because Bernoulli distribution has no parameter for # of trials.

Bernoulli random variable becomes distributed under Binomial when you have multiple independent trials, which appears to satisfy the conditions of the problem at hand.

Binomial distribution has parameters n and p.

In Binomial distribution, n is # of trials and p is probability of success in any given trial.

Therefore, X (i.e., # successes) is distributed under Binomial with n = 9 and p = 0.8.

Keep in mind, Binomial distribution measures probability of # successes in # of trials.

Therefore, we rephrase From at least 2 failures in 9 trials [with probability of a failure in any trial at 20%] To at most 7 successes in 9 trials [with probability of a success in any trial at 80%].

Solve: 1 - P(X >= 7) = 1 – (P(X = 7) + P(X = 8) + P(X = 9))

Equation immediately above will not make sense at first.

Let’s think about why the equation is set up this way.

Remember, we ultimately want to solve for probability of at least 2 failures in 9 trials.

Equation above includes P(X=7), which is the probability of exactly 7 successes in 9 trials, which means P(X=7) is also probability of exactly 2 failures. Equation immediately above also estimates P(X=8) and P(X=9), which is probability of exactly 1 and 0 failures, respectively.

Therefore, P(X = 7) + P(X = 8) + P(X = 9) estimates the probability of at most 2 failures.

But, wait, we want probability of at least 2 failures, so let’s think this through.

Do we agree that probability of at least 2 failures plus probability of at most 2 failures is 100%?

Think about that question – you’ll see that you agree.

Once you agree, it is clear 1 – (P(X = 7) + P(X = 8) + P(X = 9))  is probability of >= 2 failures.

Distribution function of P(X = k):

P(X=k) = C(n,k)*p^(k)*(1-p)^(n-k)

Solve for k =7, k = 8, and k = 9; then solve 1 – (P(X = 7) + P(X = 8) + P(X = 9)) to solve probability of at least 2 failures in 9 trials.

I'm sure you can do the combinatoric math to solve n-choose-k [where k=7,8,9] and can do the algebraic math to solve p^k*(1-p)^(n-k) , so I'll leave the rest to you.