Suppose a biased coin comes up head with a probability of 0.2 when tossed. What is the probability of achieving 0, 1, 2, 3,and 4 heads after 4 tosses?
Let probability of success, p = 0.2, and number of trials, n = 4
$P\left(0\mathrm{heads}\right)equals;f\left(0\right)equals;P\left(Xequals;0\right)equals;\left(\genfrac{}{}{0ex}{}{4}{0}\right){\left(0.2\right)}^{0}{\left(1-0.2\right)}^{\left(4-0\right)}equals;0.4096$
$P\left(1\mathrm{heads}\right)equals;f\left(1\right)equals;P\left(Xequals;1\right)equals;\left(\genfrac{}{}{0ex}{}{4}{1}\right){\left(0.2\right)}^{1}{\left(1-0.2\right)}^{\left(4-1\right)}equals;0.4096$
$P\left(2\mathrm{heads}\right)equals;f\left(2\right)equals;P\left(Xequals;2\right)equals;\left(\genfrac{}{}{0ex}{}{4}{2}\right){\left(0.2\right)}^{2}{\left(1-0.2\right)}^{\left(4-2\right)}equals;0.1536$${}$
$P\left(3\mathrm{heads}\right)equals;f\left(3\right)equals;P\left(Xequals;3\right)equals;\left(\genfrac{}{}{0ex}{}{4}{3}\right){\left(0.2\right)}^{3}{\left(1-0.2\right)}^{\left(4-3\right)}equals;0.0256$
$P\left(4\mathrm{heads}\right)equals;f\left(4\right)equals;P\left(Xequals;4\right)equals;\left(\genfrac{}{}{0ex}{}{4}{4}\right){\left(0.2\right)}^{4}{\left(1-0.2\right)}^{\left(4-4\right)}equals;0.0016$