Problem 6. The probability of the first shooter hitting the target is $p_{1}$, and the second shooter is $p_{2}$. The shooters fired simultaneously. What is the probability that:

a) both shooters hit the target;

b) only one hits;

c) at least one hits?
Solution. Let's denote the events:

$A$ - the first shooter hits; $p(A)=p_{1}$; denote $q_{1}=p(\bar{A})=$ $=1-p_{1}$

$B$ - the second shooter hits; $p(B)=p_{2}$; denote $q_{2}=p(\bar{B})=$ $=1-p_{2}$.

Then:

$A B$ - both shooters hit;

$A \bar{B}+\bar{A} B$ - one shooter hits;

$A+B$ - at least one shooter hits.

Events $A$ and $B$ are independent.

a) By formula (10): $p(A B)=p_{1} p_{2}$

b) By formulas (6) and (10):

$$
p(A \bar{B}+\bar{A} B)=p_{1}\left(1-p_{2}\right)+\left(1-p_{1}\right) p_{2}=p_{1}+p_{2}-2 p_{1} p_{2}
$$

c) By formula (12): $p(A+B)=1-q_{1} q_{2}=1-\left(1-p_{1}\right)\left(1-p_{2}\right)=$ $=p_{1}+p_{2}-p_{1} p_{2}$