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`

)?
2 questions:

ReplyDelete1- trying dbinom(3, 8, 0.5), gives the same result as dbinom(5, 8, 0.5), can you explain why??

2- Isn't this the probability of any sequence with 5 tails at least in the 8 throws, rather than that specific sequence?

Thanks