# Problem set for binomial probability distribution 1.)Use the formula for

Problem set for binomial probability distribution
1.)Use the formula for the binomial probability distribution to find the probabilities for $$\displaystyle{n}={4}.{p}={0.5}$$ and $$\displaystyle{x}={0}.{1}.{2}.{3}$$. and 4.

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Step 1
Given data:
$$\displaystyle{n}={4},{p}={0},{5}$$ and $$\displaystyle{x}={0},{1},{2},{3},{4}$$
$$\displaystyle{q}={1}-{p}$$
$$\displaystyle={1}-{0},{5}$$
$$\displaystyle={0},{5}$$
Binomial Probability distribution:
$$P(X=n)=\sum_{x=0}^{n} \left(\begin{array}{c}n\\ x\end{array}\right)p^{x}q^{n-x}$$
$$\displaystyle{P}{\left({X}-{4}\right)}={P}{\left({0}\right)}+{P}{\left({1}\right)}+{P}{\left({2}\right)}+{P}{\left({3}\right)}+{P}{\left({4}\right)}$$
$$=\left(\begin{array}{c}4\\ 0\end{array}\right)\left(0,5\right)^4\left(0,5\right)^\left\{4-0\right\}+\left(\begin{array}{c}4\\ 1\end{array}\right)\left(0,5\right)^4\left(0,5\right)^\left\{4-1\right\}+\left(\begin{array}{c}4\\ 2\end{array}\right)\left(0,5\right)^4\left(0,5\right)^\left\{4-2\right\}+\left(\begin{array}{c}4\\ 3\end{array}\right)\left(0,5\right)^4\left(0,5\right)^\left\{4-3\right\}+\left(\begin{array}{c}4\\ 4\end{array}\right)\left(0,5\right)^4\left(0,5\right)^\left\{4-4\right\}$$
$$\displaystyle{P}{\left({X}={4}\right)}={0.0039}+{0.03125}+{0.09375}+{0.125}+{0.0625}={0.3164}$$
Conclusion
Hense, the binomial probability distribution of $$\displaystyle{P}{\left({X}={4}\right)}={0.03164}$$