The enemy has three types of guns that shoot down a plane with probabilities of $0.1, 0.2$ and $0.3.$

Question

The enemy has three types of guns that shoot down a plane with probabilities of $0.1, 0.2$ and $0.3.$.
It is known that the plane we sent turned out to be shot down.
It is also known that the enemy always uses exactly one gun with different frequencies
(for example, due to the difference in the prices of shells)
the first with a frequency of $0.5$, the second $0.3$, the third $0.2$. What is the probability that this was the second gun?

Answer 1

thank for tell me that Bayes' theorem
I solved it like

• $p(A) = 0.1$
• $p(B) = 0.2$
• $p(C) = 0.3$
• $P(F|A) = 0.5$
• $P(F|B) = 0.3$
• $P(F|C) = 0.2$

using Bayes' theorem the result will be like that
$\frac{0.2 * 0.3}{(0.2 * 0.3 + 0.5 * 0.1 + 0.2 * 0.3)} = 0.3529 = 3$5.29%$ $