Solution :
[tex]$ P(A \cap B) = P(\text{coin}\ C_1 \text{ chosen and 4 heads in 10 trails})$[/tex]
[tex]$= \frac{1}{3} \times ^{10}C_4 \times 0.4^4 \times 0.6^6$[/tex]
= 0.0836
P(B)= P(coin [tex]$C_1$[/tex] chosen and 4 heads in 10 trails) + P(coin [tex]$C_2$[/tex] chosen and 4 heads in 10 trail) + P(coin [tex]$C_3$[/tex] chosen and 4 heads in 10 trail)
[tex]$=\frac{1}{3}\times ^{10}C_4 \times 0.4^4 \times 0.6^6 + \frac{1}{3}\times ^{10}C_4 \times 0.5^4 \times 0.5^6 + \frac{1}{3}\times ^{10}C_4 \times 0.2^4 \times 0.8^6$[/tex]
[tex]$=0.0836+0.0684+0.0294$[/tex]
[tex]$=0.1813$[/tex]
P(A|B) = [tex]$\frac{P(A \cap B)}{P(B)}$[/tex]
[tex]$=\frac{0.0836}{0.1813}$[/tex]
[tex]$=0.4611$[/tex]
