Answer:
Option B
Explanation:
Key Idea Use the theorem of total probability
Let E1 = Event that first ball drawn is red
E2 = Event that first ball drawn is black
A= Event that second ball drawn is red
P( E1)= $\frac{4}{10}$, $P[\frac{A}{E_{1}}]=\frac{6}{12}$
$\Rightarrow$ $P(E_{2})=\frac{6}{10}, P(\frac{A}{E_{2}})=\frac{4}{12}$
By law of total probability
$P(A_{})=P( E_{1})\times P(\frac{A}{E_{1}})+P(E_{2})\times P(\frac{A}{E_{2}})$
$=\frac{4}{10}\times \frac{6}{12}+\frac{6}{10}\times\frac{4}{12}$
$=\frac{24+24}{120}=\frac{48}{120}=\frac{2}{5}$