Question

There are 5 red balls and 2 green balls. What is the probability of picking a green ball if a ball is taken initially without replacing?

Answer 1

Let $E_1,E_2$ and $A$ be three events such that : $E_1=$ Initially taken ball is red

$E_2=$ Initially taken ball is green

$A=$ second drawn ball is green

Then $P(E_1)=\frac{5}{7},P(E_2)=\frac{2}{7}$ and $P(A/E_1)=\frac{2}{6}=\frac{1}{3},P(A/E_2)=\frac{1}{6}$

Therefore the required probability is $P(A)=P(E_1).P(A/E_1)+P(E_2).P(A/E_2)$

$=\frac{5}{7}.\frac{1}{3}+\frac{2}{7}.\frac{1}{6}=\frac{5}{21}+\frac{1}{21}=\frac{6}{21}=\frac{2}{7}$