One of the shooters hits the target with a probability of 0.7, and the other – 0.5. After each of them has shot one time

One of the shooters hits the target with a probability of 0.7, and the other – 0.5. After each of them has shot one time, there is one hole in the target. Find the probability that the second shooter has hit.

Consider the events:
A1 – the first shooter hits P (A1) = 0.7;
A2 – the second shooter hits P (A2) = 0.5;
A`1 – the first shooter missed P (A`1) = 1 – 0.7 = 0.3;
A`2 – the second shooter missed P (A`2) = 1 – 0.5 = 0.5;
Let’s put forward the following hypotheses:
H1 – the first shooter hits, the second doesn’t.
P (H1) = 0.7 * 0.5 = 0.35;
H2 – the second shooter hits, the first one doesn’t.
P (H2) = 0.3 * 0.5 = 0.15;
H3 – both shooters miss.
P (H3) = 0.3 * 0.5 = 0.15;
H3 – both shooters hit the target.
P (H4) = 0.7 * 0.5 = 0.35;
As a result of the experiment, event B was observed – the target was hit.
The conditional probabilities of this event under the hypotheses made are:
P (B | H1) = 1;
P (B | H2) = 1;
P (B | H3) = 0;
P (B | H4) = 0.
Using the Bayes formula, we find the probability of hypothesis H2 after the experiment:
P (H2 | B) = (P (H2) P (B | H2)) / (P (H1) P (B | H1) + P (H2) P (B | H2) + P (H3) P (B | H2) + P (H4) P (B | H4)) = (0.15 1) / ((0.35 1) + (0.15 1) + 0.15 0 + 0.35 0) = 0.15 / 0.5 = 0.3.
Answer: The probability of hitting the second shooter is 0.3.