 # An event has a probability p = \frac{5}{8}. Find Karen Simpson 2021-12-12 Answered
An event has a probability $p=\frac{5}{8}$. Find the complete binomial distribution for $n=6$ trials.
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Given
$p=\frac{5}{8}=0.625$
$n=6$
The binomial distribution is given by
$P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}{q}^{n-x}$
$P\left(X=x\right)=\left(\begin{array}{c}6\\ x\end{array}\right)\left(0.625{\right)}^{x}\left(1-0.625{\right)}^{6-x}$
The probability distribution table is given as under
$\begin{array}{|cc|}\hline x& p\left(x\right)\\ 0& 0.002781\\ 1& 0.027809\\ 2& 0.115871\\ 3& 0.257492\\ 4& 0.321865\\ 5& 0.214577\\ 6& 0.059605\\ \hline\end{array}$
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Step 1
P(0) Probability of exactly 0 successes
If using a calculator, you can enter and $X=0$ into a binomial probability distribution function (PDF). If doing this by hand, apply the binomial probability formula:
$P\left(X\right)=\left(\begin{array}{c}n\\ x\end{array}\right)×{p}^{x}×\left(1-p{\right)}^{n-x}$
The binomial coefficient, $\left(\begin{array}{c}n\\ x\end{array}\right)$ is defined by
$\left(\begin{array}{c}n\\ x\end{array}\right)=\frac{n!}{X!\left(n-X\right)!}$
The full binomial probability formula with the binomial coefficient is
$P\left(X\right)=\frac{n!}{X!\left(n-X\right)!}×{p}^{x}×{\left(1-p\right)}^{n-x}$
Where n is the number of trials, p is the probability if success on a single trial, and X is the number of successes. Substituting in values for this problem, and $X=0$
$P\left(0\right)=\frac{6!}{0!\left(6-0\right)!}×{0.625}^{0}×{\left(1-0.625\right)}^{6-0}$
Evaluting the expression, we have
$P\left(0\right)=0.0027809143066406$
Step 2
If we apply the binomial probability formula, or a calculator's binomial probability distribution (PDF) function, to all possible values of X for 6 trials, we can construct a complete binomial distribution table. The sum of the probabilities in this table will always be 1. The complete binomial distribution table for this problem, with $p=0.625$ and 6 trials is:
$P\left(0\right)=0.0027809143066406$
$P\left(1\right)=0.027809143066406$
$P\left(2\right)=0.11587142944336$
$P\left(3\right)=0.25749206542969$
$P\left(4\right)=0.32186508178711$
$P\left(5\right)=0.21457672119141$
$P\left(6\right)=0.059604644775391$