16.02 Binomial distributions

Find the value of $\nCr{5}{3}\cdot\left(0.8\right)^3\cdot\left(0.2\right)^2+\nCr{5}{4}\cdot\left(0.8\right)^4\cdot0.2+\nCr{5}{5}\cdot\left(0.8\right)^5\cdot\left(0.2\right)^0$5C3·(0.8)3·(0.2)2+5C4·(0.8)4·0.2+5C5·(0.8)5·(0.2)0.

$X$X is a binomial variable with the probability mass function $P$P$($($X=k$X=k$)$)$=$=$\nCr{5}{k}\cdot\left(0.7\right)^k\cdot\left(0.3\right)^{5-k}$5Ck·(0.7)k·(0.3)5−k.

Census data show that $80%$80% of the population in a particular country have brown eyes.

A random sample of $900$900 people is selected from the population.