From a group of 20 hunters, 5 hit a target with probability $\frac45$, 7 with probability $\frac35$ and 8 with probability $\frac12$. A hunter taken at random shoots, without hitting the target. Which is the probability that he belongs to the first group?
Is it okay to consider $A_i$: He is from group $i$, $B$: He hits target and calculate first $P(B)$ with law of total probability and then use Bayes's theorem?
On
By applying Bayes' theorem we obtain
$$P(A_i|B) = \frac{P(B | A_i) P(A_i)}{P(B)} = \frac{P(B | A_i) P(A_i)}{\sum_{j=1}^n P(B | A_j) P(A_j)}=\\\frac{\frac5{20}\cdot\frac15}{\frac5{20}\cdot\frac15+\frac7{20}\cdot\frac25+\frac8{20}\cdot\frac12}=\frac{\frac1{20}}{\frac1{20}+\frac7{50}+\frac15}=\frac5{5+14+20}=\frac5{39}$$
The law of total probability will be enough. $$P=\frac{\frac5{20}\cdot\frac15}{\frac5{20}\cdot\frac15+\frac7{20}\cdot\frac25+\frac8{20}\cdot\frac12}=\frac{\frac1{20}}{\frac1{20}+\frac7{50}+\frac15}=\frac5{5+14+20}=\frac5{39}$$