What is the probability, flipping a coin 8 times, to obtain the sequence HHTTTHTT? (H = head; T= tail)

The theory teaches us that to solve this question, we can simply use the following formula:

$$f(x)=P(X=x)=B(n,p)=\begin{pmatrix}n\\ x \end{pmatrix} \cdot p^x \cdot q^{n-x}=\frac{n!}{x!(n-x)!}$$

To solve a problem like this, we can use in R the function `dbinom(x, n, p)`

. The coin flipping follow a binomial distribution, in which every event can be H or T. Suppose that T is the number of successes `x`

(in this case `x = 5`

), while `n`

is the number independet experiments (in this case `n = 8`

). The probability of success is `p = 0.5`

. Put these data into R and get the answer:

dbinom(5, 8, 0.5)

[1] 0.21875

The probability of obtaining that particular sequence is equal to 21,875%.

What probability we would have obtained if we had chosen H as the success (ie by imposing `x = 3`

)?

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**Tags:** Bernoulli, Binomial distribution, Probability, R functions