Question

From a box containing 4 black balls and 2 green balls,3 balls are dawn in succession. Each ball is placed back in the box before the next draw is made. Let G be a random variable representing the number of green balls that occur. Find the values of the rabdom variable G

Answer 1

This is a binomial distribution with $n=3, \;p=\frac{2}{6}=\frac{1}{3}.$

$P(G=x)=C_3^x(\frac{1}{3})^x(\frac{2}{3})^{3-x}.$

$P(G=0)=C_3^0(\frac{1}{3})^0(\frac{2}{3})^{3}=\frac{8}{27}.$

$P(G=1)=C_3^1(\frac{1}{3})^1(\frac{2}{3})^{2}=\frac{4}{9}.$

$P(G=2)=C_3^2(\frac{1}{3})^2(\frac{2}{3})^{1}=\frac{2}{9}.$

$P(G=3)=C_3^3(\frac{1}{3})^3(\frac{2}{3})^{0}=\frac{1}{27}.$