# Statistics Basic Probability Help?

Hello Answers,

I need some help with the first part of the attached probability problem. My notes tell me the solution should be something along the lines of (2/3)*(1/3)*(1/3)=.0741, but it seems that isn't right, can anyone suggest another way to do this?

### 1 Answer

- PuzzlingLv 77 months agoFavorite Answer
You have sort of started out on the right foot, but you are only doing one order of heads and tails.

What you calculated was the probability the first coin flip was heads and the second two coin flips were tails:

P(HTT) = 2/3 * 1/3 * 1/3 = 2/27

But what you didn't account for was that there are 3 ways to pick which coin flip is heads. There are 3 flips and you choose 1 to be the head.

C(3,1) = 3 ways

Thus there are 3 ways {HTT, THT, TTH} to have exactly 1 head, so you need to multiply your answer by 3.

3 * 2/3 * 1/3 * 1/3 = 6/27 = 2/9

Answer:

P(3 flips, exactly 1 head) = 2/9 (or ~0.2222)

The others are basically the same with slight variations. The second one switches to counting tails (n = 5, k = 2, p = 1/3, q = 2/3). The third wants *at least* 3 tails.

What this lesson covers is the binomial probability formula. You have two outcomes (success/failure) --> binomial. You have multiple trials (flips) and each is independent but with a specific probability of success --> P(heads) = 2/3.

The general formula for exactly k successes is:

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

n : number of trials (e.g. 5 flips)

k : number of successes (e.g. exactly 2 *tails*)

p : probability of success (p = 1/3) <-- tails

q : probability of failure (q = 1-p = 2/3) <-- heads

C(n,k) <-- number of ways from n items to choose a combination of k items = n! / (k! (n-k)!)

P(X = 2) = C(5,2) * (1/3)^2 * (2/3)^3

= 10 * 1/9 * 8/27

= 80/243

≈ 0.3292

I think you can figure out the last one. You'll have to do *two* calculations because it says "at least 3 tails". First calculate the probability of exactly 3 tails. Then repeat the process for exactly 4 tails. Add the two together.

P(X ≥ 3) = P(X = 3) + P(X = 4)

Sorry, slight correction to the formula --> P(X = k) = C(n,k) * p^k * q^(n-k)