Solve probability problem using Bayes theorem

Question

Problem

There is a group of 20 soldiers. 4 of them are best shooters, 10 are good and 6 are bad. The chance of hitting the target for best shooter $= 0.9$, for good = $0.7$ and for bad = $0.5$. Commander picks 2 random shooters from group and they take a shot. Find the probability that the target was shot at least once ($\ge 1$).

My steps:

I understand how to solve this problem without using Bayes theorem.

Target hitting rates:

$A|H_{1} = 0.9$

$A|H_{2} = 0.7$

$A|H_{3} = 0.5$

Probability of picking:

$P(H_{1}) = \frac{4}{20}$

$P(H_{2}) = \frac{10}{20}$

$P(H_{3}) = \frac{6}{20}$

So the possibility of hitting a target at least once for 2 shooters is calculated like this:

$$P(A)=2\cdot(P(A|H_{1})\cdot P(H_{1})+P(A|H_{2})\cdot P(H_{2})+P(A|H_{3})\cdot P(H_{3}))$$

$$P(A)=2\cdot (0.9 \cdot 0.2+0.7\cdot 0.5+ 0.5\cdot 0.3)=1.36$$

Question

How to solve this using Bayes theorem?

Answer 1

Issues with your "not using Bayes' rule" solution: $\def\P{\operatorname{\sf P}}$

1. You are correctly using the Law of Total Probability.

   $$\P(A_1)~=~ {{\P(A_1\mid H_{1,1})\P(H_{1,1})}+{\P(A_1\mid H_{1,2})\P(H_{1,2})}+{\P(A_1\mid H_{1,3})\P(H_{1,3})}}$$

   Where I am using $A_1,A_2$ to denote the events of the first and second shooter hitting the target).   Also you are correct that $\P(A_1)~=~\P(A_2)$.

2. You appear to be trying to say $\P(A_1\cup A_2) = \P(A_1) + \P(A_2)$ , however, this is not so.   These events are not disjoint.

   The additive rule is that $\P(A_1\cup A_2) = \P(A_1)+\P(A_2)-\P(A_1\cap A_2)$ and the last term is not zero.   This is why you are obtaining an over certain probability.

3. To obtain the joint probability, $\P(A_1\cap A_2)$, do similar, but remember that the events enumerating which categories the two shooters belong to will not be independent.

   $$\begin{align}\P(A_1\cap A_2)~=&~ {{\P(A_1\mid H_{1,1})\big({\P(A_2\mid H_{2,1})\P(H_{1,1},H_{2,1})}+{\P(A_2\mid H_{2,2})\P(H_{1,1},H_{2,2})}+{\P(A_2\mid H_{2,3})\P(H_{1,1},H_{2,3})}\big)}}+{{\P(A_1\mid H_{1,2})\big({\P(A_2\mid H_{2,1})\P(H_{1,2},H_{2,1})}+{\P(A_2\mid H_{2,2})\P(H_{1,2},H_{2,2})}+{\P(A_2\mid H_{2,3})\P(H_{1,2},H_{2,3})}\big)}}+{{\P(A_1\mid H_{1,3})\big({\P(A_2\mid H_{2,1})\P(H_{1,3},H_{2,1})}+{\P(A_2\mid H_{2,2})\P(H_{1,3},H_{2,2})}+{\P(A_2\mid H_{2,3})\P(H_{1,3},H_{2,3})}\big)}}\end{align}$$

To use Bayes' rule. ... Why do you even think you would use Bayes' rule?