# What is the probability of X= 6 successes, using the binomial formula?

## Let $n = 13 , p = .43$.

Jan 7, 2017

$P \left(X = 6\right) \approx 0.2121$, or approximately 21.21%.

#### Explanation:

The binomial formula states that the probability of getting exactly $x$ successes out of $n$ trials (where each independent trial has success probability $p$) is

$P \left(X = x\right) = \left(\begin{matrix}n \\ x\end{matrix}\right) {p}^{x} {\left(1 - p\right)}^{n - x} , \text{ } x = 0 , 1 , 2 , \ldots , n$

where

• $\left(\begin{matrix}n \\ x\end{matrix}\right)$ is the number of ways to position $x$ successes in a sequence of $n$ trials, equal to (n!)/(x!(n-x)!);
• ${p}^{x}$ is the probability of getting those $x$ independent successes; and
• ${\left(1 - p\right)}^{n - x}$ is the probability of failure for the remaining $n - x$ independent trials.

To obtain an answer, we simply plug in the given values of $n$, $p$, and $x$:

$P \left(X = 6\right) = \left(\begin{matrix}13 \\ 6\end{matrix}\right) {\left(.43\right)}^{6} {\left(1 - .43\right)}^{13 - 6}$

color(white)(P(X=6))=(13!)/(6!(13-6)!)(.43)^6(.57)^7

$\textcolor{w h i t e}{P \left(X = 6\right)} \approx \left(1716\right) \left(0.006321\right) \left(0.015949\right)$

$\textcolor{w h i t e}{P \left(X = 6\right)} \approx 0.2121$

So, out of 13 trials, the probability of obtaining exactly 6 successes is

$P \left(X = 6\right) \approx 0.2121$
color(white)(P(X=6)) = 21.21%.